M. J. Roberts - 8/16/04
Chapter 5 - The Fourier Transform
Selected Solutions
(In this solution manual, the symbol, ⊗, is used for periodic convolution because the
preferred symbol which appears in the text is not in the font selection of the word
processor used to create this manual.)
1. The transition from the CTFS to the CTFT is illustrated by the signal,
 t
 t 1
x( t) = rect   ∗ comb 
 w  T0
 T0 
or
x( t) =
∞
 t − nT0 
 .
w 
∑ rect 
n =−∞
The complex CTFS for this signal is given by
X[ k ] =
Plot the “modified” CTFS,
 kw 
Aw
sinc  .
T0
 T0 
T0 X[ k ] = Aw sinc( w ( kf 0 )) ,
for w = 1 and f 0 = 0.5, 0.1 and 0.02 versus kf 0 for the range −8 < kf 0 < 8 .
kg
and is a function of spatial position, x, in
m3
meters. Write the mathematical expression for its CTFT, M( y ) . What are the units of
2. Suppose a function, m( x ) , has units of
M and y?
M( y ) =
∞
∫ m( x )e
−∞
5-1
− j 2πyx
dx
M. J. Roberts - 8/16/04
kg
because of the multiplication by dx in the integral and the units of
m2
y are m−1 because they are always the reciprocal of the units of the independent variable of
the function transformed.
The units of M are
3 . Using the integral definition of the Fourier transform, find the CTFT of these
functions.
(a) x( t) = tri( t)
Substitute the definition of the triangle function into the integral and use even and
odd symmetry to reduce the work.
1
Also, use sin ( x ) sin ( y ) =  cos ( x − y ) − cos ( x + y )  to put the final expression into
2
the form of a sinc-squared function.
 1
 1
(b) x( t) = δ  t +  − δ  t − 
 2
 2
Use the sampling property of the impulse.
4. In Figure E4 there is one example each of a lowpass, highpass, bandpass and bandstop
signal. Identify them.
x(t)
x(t)
(a)
(b)
t
x(t)
t
x(t)
(c)
t
(d)
t
Figure E4 Signals with different frequency content
(a)
bandstop
(b)
bandpass
Composed of very high and very low frequencies and nothing
between
Looks most like a sinusoid.
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M. J. Roberts - 8/16/04
(c)
(d)
lowpass
highpass
Smoother than the others, therefore has mostly low frequencies.
Fast variation without any underlying low frequencies
5 . Starting with the definition of the CTFT find the radian-frequency form of the
generalized CTFT of a constant. Then verify that a change of variable, ω → 2πf ,
yields the correct result in cyclic-frequency form. Check your answer against the
Fourier transform table in Appendix E.
Similar to the derivation in the text for cyclic frequencies. Use the scaling property of
the impulse to compare with the cyclic-frequency result.
6. Starting with the definition of the CTFT, find the generalized CTFT of a sine of the
form, A sin(ω 0 t) and check your answer against the results given above. Check your
answer against the Fourier transform table in Appendix E.
Similar to Exercise 5.
7. Find the CTFS and CTFT of each of these periodic signals and compare the results.
After finding the transforms, formulate a general method of converting between the
two forms for periodic signals.
(a)
x( t) = A cos(2πf 0 t)
The CTFS is simply two impulses, X[ k ] =
The CTFT is X( f ) =
A
(δ[k − 1] + δ[k + 1]) .
2
A
(δ ( f − f 0 ) + δ ( f + f 0 )) = X[1]δ ( f − f 0 ) + X[−1]δ ( f + f 0 ) .
2
X( f ) =
∞
∑ X[k ]δ ( f − kf )
k =−∞
0
The CTFT is a set of continuous-frequency impulses whose weights at frequencies,
kf0 , are the same as the weights of the discrete-harmonic-number impulses at harmonic
number, k, in the CTFS harmonic function.
(b)
x( t) = comb( t)
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M. J. Roberts - 8/16/04
8. Let a signal be defined by
x( t) = 2 cos( 4πt) + 5 cos(15πt) .
1
1


Find the CTFT’s of x t −  and x t +  and identify the resultant phase shift of each
 40 
 20 
sinusoid in each case. Plot the phase of the CTFT and draw a straight line through the 4
phase points which result in each case. What is the general relationship between the slope
of that line and the time delay?
The slope of the line is −2πf times the delay.
9. Using the frequency-shifting property, find and plot versus time the inverse CTFT of
 f + 20 
 f − 20 
X( f ) = rect 
 .
 + rect 
 2 
 2 
10. Find the CTFT of
x( t) = sinc( t) .
Then make the transformation, t → 2 t , in x( t) and find the CTFT of the transformed
signal.
11. Using the multiplication-convolution duality of the CTFT, find an expression for y( t)
which does not use the convolution operator, ∗ , and plot y( t) .
(a)
y( t) = rect ( t) ∗ cos(πt)

1 
1
1  

y ( t ) = F −1 sinc ( f ) δ  f −  + δ  f +   

2 
2
2  

y (t ) =
1 −1  
1
1
 1 
 1

F δ  f −  sinc   + δ  f +  sinc  −  
 2 
 2

2
2
2
 
Using the equivalence property of the impulse,
y (t ) =
1 −1  2 
1 2 
1  1
1
1 
 

F  δ  f −  + δ  f +   = F −1 δ  f −  + δ  f +  

2
2 π 
2  π
2
2 
π 
 
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M. J. Roberts - 8/16/04
y( t) =
(b)
Similar to (a)
(c)
 t
y( t) = sinc( t) ∗ sinc 
 2
2
cos(πt)
π
This convolution would be very difficult to do directly in the time domain.
But, using transform methods, it is quite easy.
t
y ( t ) = F −1 {rect ( f ) × 2 rect ( 2 f )} = 2F −1 {rect ( 2 f )} = sinc  
 2
(d)
Similar to (c).
(e)
y( t) = e − t u( t) ∗ sin(2πt)
Use the equivalence property of the impulse, then find a common
denominator and simplify. Then, use

 B
A cos( x ) + B sin( x ) = A 2 + B 2 cos x − tan −1   
 A 

to get
y (t ) =
cos ( 2π t + 2.984 )
1 + ( 2π )
2
12. Using the CTFT of the rectangle function and the differentiation property of the CTFT
find the Fourier transform of
x( t) = δ ( t − 1) − δ ( t + 1) .
Check your answer against the CTFT found using the table and the time-shifting property.
 t
Let y( t) = − rect   .
 2
Then x( t) =
d
(y(t)) .
dt
(This comes from the definition of a
generalized derivative in Chapter 2.)
t F
− rect   ←
→ −2 sinc ( 2 f )
 2
Using the differentiation property of the CTFT,
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M. J. Roberts - 8/16/04
d
 t  F
− rect    ←
→ j 2π f  −2 sinc ( 2 f )  = − j 4π f sinc ( 2 f )

 2
dt 
sin ( 2π f )
d
 t  F
− rect    ←
→ − j 4π f
= − j 2 sin ( 2π f )

 2
2π f
dt 
Use the the CTFT of the impulse and the time-shifting property to check this answer.
13. Find the CTFS and CTFT of these periodic functions and compare answers.
(a)
1
 t
x( t) = rect ( t) ∗ comb 
 2
2
Find the CTFS harmonic function using the integral definition or Appendix E.
X[ k ] =
1
 k
sinc  
 2
2
∞
1
k

X( f ) = sinc( f ) comb(2 f ) = sinc( f ) ∑ δ  f − 

2
2
k =−∞
∞
∞
k
1
 k 
X( f ) = ∑ sinc  δ  f −  = ∑ X[ k ]δ ( f − kf 0 )
 2 
2  k =−∞
k =−∞ 2
The CTFT impulses at kf0 have the same strengths as the CTFT harmonic
function impulses at k.
(b)
x( t) = tri(10 t) ∗ 4 comb( 4 t)
Find the CTFS harmonic function using the integral definition or Appendix E.
4 
cos  π k  − 1
5 
2
 2k  5
X [ k ] = sinc 2   =
 5 4
5
( π k )2
X( f ) =
∞
1
 f
 f 4
 f
sinc 2   comb  = sinc 2   ∑ δ ( f − 4 k )
 10 
 4  10
 10  k =−∞
10
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M. J. Roberts - 8/16/04
2 ∞
 2k 
X( f ) = ∑ sinc 2   δ ( f − 4 k )
 5
5 k =−∞
Checks with CTFS.
14. Using Parseval’s theorem, find the signal energy of these signals.
(a)
 t
x( t) = 4 sinc 
 5
(b)
x( t) = 2 sinc 2 ( 3t)
∞
Ex =
∫ x(t)
2
∞
dt =
−∞
∫ X( f )
2
∞
df =
−∞
∫
−∞
2
∞
2  f
4
 f
tri  df = ∫ tri2   df
3  3
9 −∞  3 
3
3
3
f
f 2
8
8 
8  2f
2 f 
E x = ∫ tri   df = ∫ 1 −  df = ∫ 1 −
+  df
3
9
9 0  3
9 0
3
9 0
2
3
f 2 f 3  8  9 27  8
8
Ex =  f −
+
= 3− +  =
3 27  9
9
3 27  0 9 
 t − 8
15. What is the total area under the function, g( t) = 100 sinc
 ?
 30 
∞
Use
∫ g(t)dt = G(0)
−∞
16. Using the integration property, find the CTFT of these functions and compare with the
CTFT found using other properties.
(a)
1
, t <1

g( t) = 2 − t , 1 < t < 2
0
, elsewhere

Find the CTFT of the derivative of this function (which is two separated
rectangles). Then use the integration property to find the CTFT of the
original function.
(b)
 t
g( t) = 8 rect  
 3
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M. J. Roberts - 8/16/04
17. Sketch the magnitudes and phases of the CTFT’s of these signals in the f form.
(a)
x (t ) = δ ( t − 2 )
Remember, there are many alternate correct ways of plotting phase. So
your phase plot may be correct even if it does not look like the answer
provided in the text.
(b)
x ( t ) = u ( t ) − u ( t − 1)
This can be done directly using the two unit steps or by converting them
into a shifted rectangle.
(c)
 t + 2
x( t) = 5 rect 

 4 
(d)
x( t) = 25 sinc(10( t − 2))
(e)
x( t) = 6 sin(200πt)
(f)
x( t) = 2e −3 t u( 3t)
(g)
x (t ) = 4 e
−3t
2
= 4e
 3 
−π 
t
 π 
2
π − π3
←→ X ( f ) = 4
e
3
2
F
f2
18. Sketch the magnitudes and phases of the CTFT’s of these signals in the ω form.
∞
1
t F
ω
comb   ←
→ X ( jω ) = comb   = π ∑ δ (ω − kπ )
 2
π
2
k = −∞
(a)
x (t ) =
(b)
x( t) = sgn(2 t)
(c)
(e)
(f)
 t − 4
x( t) = 10 tri

 20 
π

cos  200π t − 

4
x (t ) =
4
x( t) = 2e −3 t u( t)
 t + 1
sinc 2 

 3 
(d)
x( t) =
10
1 


cos  200π  t −



800  
=
4
(g)
19. Sketch the inverse CTFT’s of these functions.
5-8
x( t) = 7e −5 t
M. J. Roberts - 8/16/04
(a)
 f
X( f ) = −15 rect  
 4
(c)
F
e− a t ←
→
2a
a + ( 2π f )
2
X( f ) =
(b)
2
, f →
F
x ( t ) = 6π e−3 2 π t ←
→ X( f ) =
sinc(−10 f )
30
f
, x( t) → 2π x(2πt)
2π
18
9+ f2
δ ( f − 3) + δ ( f + 3)
6
(d)
X( f ) =
1
10 + jf
(e)
X( f ) =
(f)
X( f ) = 8δ (5 f )
(g)
X( f ) = −
3
jπf
20. Sketch the inverse CTFT’s of these functions.
(a)
e
−π t 2
F
←→ e
−
ω2
4π
2
t
2
1 − 16
F
x (t ) =
e ←
→ X ( jω ) = e−4 ω
4 π
(b)
(d)
(e)
1  t 
x

4 π 4 π
( 4 π ω )2
ω2
−16 π
−
= e 4π = e 4π
, ω → 4 πω , x( t) →
ω
X( jω ) = 7 sinc 2  
π
(c)
X( jω ) = jπ [δ (ω + 10π ) − δ (ω − 10π )]
 4ω 
comb 
 π 
X( jω ) =
5
F
sgn ( t ) ←
→
x (t ) =
2
jω
F
, 1←
→ 2πδ (ω )
5π
5π
F
sgn ( t ) + 5 ←
→ X ( jω ) =
+ 10πδ (ω )
jω
2
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M. J. Roberts - 8/16/04
x(t)
18
-1
(f)
X( jω ) =
1
-4
6
3 + jω
(g)
t
X( jω ) = 20 tri(8ω )
21. Find the CTFT’s of these signals in either the f or ω form, whichever is more
convenient.
(a)
x( t) = 3 cos(10 t) + 4 sin(10 t)
ω 0 = 10 and f0 =
5
Therefore the ω form is slightly more convenient.
π
X ( jω ) = 3π δ (ω − 10 ) + δ (ω + 10 )  + j 4π δ (ω + 10 ) − δ (ω − 10
0 ) 
X ( jω ) = ( 3 − j 4 ) πδ (ω − 10 ) + ( 3 + j 4 ) πδ (ω + 10 )
X( jω ) = (5πe − j 0.927 )δ (ω − 10) + (5πe j 0.927 )δ (ω + 10)
(b)
 t − 1
 t
x( t) = comb  − comb

 2 
 2
(c)
 
x( t) = 4 sinc( 4 t) − 2 sinc 4  t −
 
(d)
x( t) = 2e( −1+ j 2π ) t + 2e( −1− j 2π ) t u( t)
[
1
 
  − 2 sinc 4  t +

 
4 
]
(e)
1

4
x( t) = 4 e
−
t
16
22. Sketch the magnitudes and phases of these functions. Sketch the inverse CTFT’s of
the functions also.
(a)
X( jω ) =
10
4
−
3 + jω 5 + jω
(b)
5-10

 f + 1 
 f − 1
X( f ) = 4 sinc

 + sinc
 2  
 2 

M. J. Roberts - 8/16/04
(c)
F
x ( t ) = 1.6 sinc 2 ( 8t ) sin ( 4π t ) ←
→ X( f ) =
j   f + 2
 f − 2 
− tri 
tri 


 8  
10   8 
|X( f )|
x(t)
0.1
0.5
-0.5
0.5
-15
t
15
f
15
f
Phase of X( f )
π
-0.5
-15
-π
(d)
X( f ) = δ ( f + 1050) + δ ( f + 950) + δ ( f − 950) + δ ( f − 1050)
(e)
δ ( f + 1050) + 2δ ( f + 1000) + δ ( f + 950) 
X( f ) = 

+δ ( f − 950) + 2δ ( f − 1000) + δ ( f − 1050) 
23. Sketch these signals versus time. Sketch the magnitudes and phase of their CTFT’s in
either the f or ω form, whichever is more convenient.
(a)
 1
x( t) = rect (2 t) ∗ comb( t) − rect (2 t) ∗ comb t − 
 2
1
 f
X( f ) = sinc  comb( f )(1 − e − jπf )
 2
2
X( f ) = je
X( f ) =
−j
∞
πf
2
∑e
 πf 
 f
sinc  comb( f ) sin 
 2
 2
π
− j ( k −1)
2
k =−∞
 k   πk 
sinc  sin  δ ( f − k )
 2  2 
Non-zero only for odd values of k. At those odd values, e
always evaluates to +1. Therefore
X( f ) =
∞
 k
∑ sinc 2  δ ( f − k )
k =−∞
k ≠0
5-11
π
− j ( k −1)
2
 πk 
sin  ,
 2
M. J. Roberts - 8/16/04
(b)
Same as answer in part (a).
(c)
x( t) = e
−
t
4
u( t) ∗ sin(2πt)
Find transform, use impulse equivalence property, get a common
denominator and simplify.
x( t) =
(d)
4 sin(2πt) − 32π cos(2πt)
1 + 64π 2
x( t) = e −πt ∗ [rect (2 t) ∗ comb( t)]
2
This is a “Gaussian” smoothing operation. The square wave is heavily
smoothed. So heavily, in fact, that about all that remains is the average
value (1/2) plus a small sinusoid.
(e)
x( t) = rect ( t) ∗ [tri(2 t) ∗ comb( t)]
x ( t ) is a constant, 1/2. Can you show that by convolving directly?
(f)
x( t) = sinc(2.01t) ∗ comb( t)
Parts (f) and (g) look almost identical, yet the results are quite different.
Why? (Hint: The operation of convolving with a sinc function produces
an effect commonly known as an ideal lowpass filter. One which makes a
very fast transition in the frequency domain from passing to stopping a
signal.)
(g)
x( t) = sinc(1.99 t) ∗ comb( t)
(h)
x( t) = e − t ∗ e − t
2
2
A Gaussian convolved with Gaussian produces a Gaussian.
24. Sketch the magnitudes and phases of these functions. Sketch the inverse CTFT’s of
the functions also.
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M. J. Roberts - 8/16/04
(a)
 f 
X( f ) = sinc
 ∗ δ ( f − 1000) + δ ( f + 1000)]
 100  [
The time-domain function is a “burst” of a sinusoid.
(b)
X( f ) = sinc(10 f ) ∗ comb( f )
25. Sketch these signals versus time. Sketch the magnitudes and phases of the CTFT’s of
these signals in either the f or ω form, whichever is more convenient. In some cases
the time sketch may be conveniently done first. In other cases it may be more
convenient to do the time sketch after the CTFT has been found, by finding the inverse
CTFT.
(a)
x( t) = e −πt sin(20πt)
(b)
x( t) = cos( 400πt) comb(100 t) =
2
1 ∞
n 

cos( 4πn )δ  t −

∑
 100 
100 n −−∞
A graph of this function looks just like a comb function, even though the
comb is multiplied by a cosine. Why?
Given that the time-domain function looks like a comb function you should
expect its CTFT to look like the CTFT of a comb, which is another comb.
1
1
 f 
δ ( f − 200) + δ ( f + 200)] ∗
comb

[
 100 
100
2




1 
 f + 200  
 f − 200 
X( f ) =

 + comb
comb



200 144
2100
443 1442100
443 
 f 
 f 


= comb 
= comb 


 100 
 100 


X( f ) =
X( f ) =
∞
1
1 ∞  f
 f 

comb
δ
−
k
=
δ ( f − 100 k )
=


 100  100 k∑
 100  k∑
100
=−∞
=−∞
(c)
x( t) = [1 + cos( 400πt)] cos( 4000πt)
(d)
x( t) = [1 + rect (100 t) ∗ 50 comb(50 t)] cos(500πt)
5-13
M. J. Roberts - 8/16/04
(e)
 t
x( t) = rect   comb( t)
 7
X( f ) = 7 sinc( 7 f ) ∗ comb( f ) = 7 sinc( 7 f ) ∗
∞
∞
∑ δ ( f − k)
k =−∞
X( f ) = 7 ∑ sinc( 7( f − k ))
k =−∞
This periodically-repeated sinc function is equivalent to a Dirichlet
function.
26. Sketch the magnitudes and phases of these functions. Sketch the inverse CTFT’s of
the functions also.
(a)
 f
X( f ) = sinc  comb( f )
 4
(b)

 f + 1 
 f − 1
X( f ) = sinc
 comb( f )
 + sinc
 4  
 4 

(c)
X( f ) = sinc( f ) sinc(2 f )
1
 t 1
 t
x( t) = rect ( t) ∗ rect   = rect ( t) ∗ rect  


 2
2
2
2
The result of the graphical convolution can be expressed in the form,
x( t) =
1   2t 

3 tri  − tri(2 t)  . A generalization of this result his leads to the

4   3

pair,
a + b  2t  a − b  2t  F
tri 
tri 
←→ ab sinc ( af ) sinc ( bf )
−
 a − b 
 a + b 
2
2
a>b>0
in Appendix E.
27. Sketch these signals versus time and the magnitudes and phases of their CTFT’s.
(a)
x( t) =
d
[sinc(t)]
dt
(b)
5-14
x( t) =
d
 t
4 rect   

 6 
dt 
M. J. Roberts - 8/16/04
x (t ) =
(c)
d
 tri ( 2t ) ∗ comb ( t ) 
dt 
Using the convolution property that the derivative of a convolution is the
convolution of either of the two functions with the derivative of the other,
 

1
1 
 
 
x ( t ) = 2   rect  2  t +   − rect  2  t −    ∗ comb ( t ) 
 
 
4
4 
 

X( f ) = j 2πf
∞
1
 f
 k
sinc 2   comb( f ) = jπ ∑ k sinc 2   δ ( f − k )
 2
 2
2
k =−∞
|X( f )|
x(t)
2
2
-2
2
-8
t
8
f
8
f
Phase of X( f )
π
-2
-8
-π
28. Sketch these signals versus time and the magnitudes and phases of their CTFT’s.
x( t) =
(a)
t
∫ sin(2πλ )dλ
−∞
X( f ) =
δ ( f + 1) − δ ( f − 1)
j
1
1
× δ ( f + 1) − δ ( f − 1)  + F ( sin ( 2π t ))  f = 0 δ ( f ) =
j 2π f 2
4π f
2
X( f ) = −
(b)
(c)
x( t) =
x( t) =
δ ( f + 1) δ ( f − 1)
1 1
−
=−
× [δ ( f + 1) + δ ( f − 1)]
4π
4π
2π 2
t
∫ rect (λ )dλ
−∞
t
∫ 3sinc(2λ )dλ
−∞
3
Let u = 2πλ . Then x( t) =
2π
2πt
3
 u
∫−∞ sinc π  du = 2π
For t ≤ 0 :
5-15
sin( u)
du
u
−∞
2πt
∫
M. J. Roberts - 8/16/04
x( t) =
0
0
2πt
−∞
sin( u)  3 
sin( u)
sin( u) 
3  sin( u)
du 
du
du
du
−
+
=
−
∫

 ∫
∫
∫
u
u
2π −∞ u
2πt
0
 2π  0 u




∞
2πt
sin( u)  3 π
3  sin(λ )
 3 π

+ Si(2πt)  =
du  =
− Si(−2πt) 
x( t) =
dλ + ∫
∫


u
2π  0 λ
2π  2
 2π  2

0 4243 
14243 1
=Si( 2πt )

 =Si( ∞ ) = π

2

“Si” is called the sine integral. It is a special function of calculus and can
be computed by the MATLAB function, sinint.
For t ≥ 0 :
−∞
0
2πt
2πt
sin( u)  3 
sin( u)
sin( u) 
3  sin( u)
du 
du + ∫
du  =
du + ∫
x( t) =
∫
− ∫
u
u
2π −∞ u
 2π  0 u

0
0
x( t) =
3
2π
π

 2 + Si(2πt) 
Therefore, for any t, x( t) =
X( f ) =
3
2π
π

 2 + Si(2πt)  .
1 3
3
 f 1
 f 3
rect   + F  3 sinc ( 2t )  f = 0 δ ( f ) =
rect   + δ ( f )
 2 2
 2 4
j 2π f 2
j 4π f
|X( f )|
x(t)
1
2
-2
-4
4
t
-1
f
2
f
π
-2
29. From the definition, find the DTFT of
x[ n ] = 10 rect 4 [ n ] .
5-16
2
Phase of X( f )
-π
M. J. Roberts - 8/16/04
and compare with the Fourier transform table in Appendix E.
Apply the definition and put into closed form by using the formula for the
summation of a geometric series,
1

∑ r = 1 − r N
n=0

 1− r
,
N −1
r =1
n
.
, otherwise
Then convert to a sine function by factoring out the proper complex exponential
and recognize the ratio of two sine functions as a Dirichlet function. Check your
answer against Appendix E.
30. From the definition, derive a general expression for the F and Ω forms of the DTFT of
functions of the form,
x[ n ] = A sin(2πF0 n ) = A sin(Ω0 n ) .
(It should remind you of the CTFT of x( t) = A sin(2πf 0 t) = A sin(ω 0 t) .) Compare with the
Fourier transform table in Appendix E.
X( F ) =
∞
∑ x[n]e
n =−∞
− j 2πFn
=
∞
∑ A sin(2πF n)e
0
n =−∞
X( F ) =
− j 2πFn
e j 2πF0 n − e − j 2πF0 n − j 2πFn
e
=A∑
j2
n =−∞
∞
A ∞
∑ e j 2π ( F0 − F ) n − e − j 2π ( F0 + F ) n
j 2 n =−∞
[
Then, using
∞
∑e
n =−∞
j 2πxn
]
= comb( x )
we get
X( F ) =
[
]
[
]
j
A
comb( F0 − F ) − comb(− F0 − F ) = A − comb( F0 − F ) + comb(− F0 − F )
j2
2
j
X( F ) = A comb( F + F0 ) − comb( F − F0 )
2
[
]
The Ω form can be found by the transformation, F →
5-17
Ω
.
2π
M. J. Roberts - 8/16/04
j
 Ω Ω0 
 Ω Ω0  
X( jΩ) = A comb
+  − comb
− 
 2π 2π 
 2π 2π  
2
[
]
X( jΩ) = jπA comb(Ω + Ω0 ) − comb(Ω − Ω0 )
31. A DT signal is defined by
 n
x[ n ] = sinc  .
 8
Sketch the magnitude and phase of the DTFT of x[ n − 2] .
32. A DT signal is defined by
 πn 
x[ n ] = sin  .
 6
Sketch the magnitude and phase of the DTFT of x[ n − 3] and x[ n + 12].
The DTFT of x[ n + 12] should be exactly the same as the DTFT of x [ n ] . Why?
33. The DTFT of a DT signal is defined by
Sketch x[ n ] .

π 
π  
2
2
 Ω
X( jΩ) = 4 rect   Ω −   + rect   Ω +    ∗ comb  .
 2π 
π 
π 
2 
2  

Start with
 n F
 wΩ 
 Ω
→ wrect 
∗ comb  
sinc   ←

 w
 2π 
 2π 
and apply the frequency-shifting and linearity properties to produce (after simplification)

π  
π 
2
2
 Ω
 n
π  F
2 sinc   cos  n  ←
→ 4  rect   Ω −   + rect   Ω +    ∗ comb
b 
 2π 
 4
2 
π 
π 
2
2 

Remember in applying the frequency-shifting property, if either (but not both) of two
functions being convolved shifts, the result of the convolution shifts by the same amount.
5-18
M. J. Roberts - 8/16/04
34. Sketch the magnitude and phase of the DTFT of
 2πn 
x[ n ] = rect 4 [ n ] ∗ cos
 .
 6 
Then sketch x[ n ] .
From the table,
F
rect Nw [ n ] ←
→ ( 2 N w + 1) drcl ( F, 2 N w + 1)
and
1
F
cos ( 2π F0 n ) ←
→  comb ( F − F0 ) + comb ( F + F0 ) 
2
X ( F ) = 9 drcl ( F, 9 ) ×
1
1
1 


comb  F −  + comb  F +  



2
6
6 
Since both functions are periodic with period, one, at every impulse in the comb function
the value of the Dirichlet function will be the same.
X(F ) =
9
1
1 

 1 
1 

drcl  , 9  comb  F −  + drcl  − , 9  comb  F +  

 6 


6 
2
6
6 
X(F ) =
1
1 
9
1 


drcl  , 9   comb  F −  + comb  F +  

6 

6
6 
2
 3π 
sin  
 2 
π
9 sin  
 6
1
1 



X ( F ) = −  comb  F −  + comb  F +  


6
6 

Then, using
1
F
cos ( 2π F0 n ) ←
→  comb ( F − F0 ) + comb ( F + F0 ) 
2
1
1 


 2π n  F

−2 cos 
←→ −  comb  F −  + comb  F +  


 6 

6
6 

and, therefore,
5-19
M. J. Roberts - 8/16/04
 2πn 
x[ n ] = −2 cos

 6 
35. Sketch the inverse DTFT of
X( F ) = [rect ( 4 F ) ∗ comb( F )] ⊗ comb(2 F ) .
Find the individual inverse DTFT’s and multiply in the time domain.
36. Using the differencing property of the DTFT and the transform pair,
 n F
tri   ←
→ 1 + cos ( 2π F ) ,
 2
1
(δ[n + 1] + δ[n] − δ[n − 1] − δ (n − 2)) . Compare it with Fourier
2
transform found using the table in Appendix E.
find the DTFT of
1
 n
The first backward difference of tri  is (δ [ n + 1] + δ [ n ] − δ [ n − 1] − δ ( n − 2)) .
 2
2
Apply the differencing property and simplify.
Other route to the DTFT:
(
1
1
F
δ [ n + 1] + δ [ n ] − δ [ n − 1] − δ [ n − 2 ]) ←
→ e j 2 π F + 1 − e− j 2 π F − e− j 4 π F
(
2
2
)
37. Using Parseval’s theorem, find the signal energy of
 n   2πn 
x[ n ] = sinc  sin
 .
 10   4 
Ex =
∞
∑ x[n]
n =−∞
2
= ∫ X( F ) dF
2
1
Find the individual DTFT’s, periodically convolve them in F and integrate the square of
the magnitude of that result over one period (one). Remember, periodic convolution of
two periodic functions is the same as the aperiodic convolution of one period of either
function with the entire other function.
5-20
M. J. Roberts - 8/16/04
Ex = ∫
1

1


j 5 rect (10 F ) ∗ comb F +  − rect (10 F ) ∗ comb F −



4

2
1 
 dF
4  
Since we are integrating only over a range of one, only one impulse in each comb is
significant.
2

1 
1
 
 
E x = 25 ∫ rect 10 F +   − rect 10 F −    dF
1
 
 
4 
4

The square of the sum equals the sum of the squares because there is no cross product; the
two rectangles do not overlap.
Ex = 5
38. Sketch the magnitude and phase of the CTFT of
x1 ( t) = rect ( t)
and of the CTFS of
1
 t
x 2 ( t) = rect ( t) ∗ comb  .
 8
8
For comparison purposes, sketch X1 ( f ) versus f and T0 X 2 [ k ] versus kf 0 on the same set
1
of axes. ( T0 is the period of x 2 ( t) and T0 = .)
f0
X1 ( f ) = sinc( f )
Using the relationship between an the CTFT of an aperiodic signal and the CTFS of a
periodic extension of that signal,
1
 k
X 2 [ k ] = f s X1 ( f sk ) = sinc 
 8
8
5-21
M. J. Roberts - 8/16/04
|T0 X2[ k]|
|X1( f )|
1
1
-4
4
f
-4
4
kf0
4
kf0
Phase of T0 X2[ k]|
Phase of X1( f )
π
π
-4
4
-4
f
-π
-π
39. Sketch the magnitude and phase of the CTFT of
x1 ( t) = 4 cos( 4πt)
and of the DTFT of
x 2 [ n ] = x1 ( nTs )
1
. For comparison purposes sketch X1 ( f ) and Ts X 2 (Ts f ) versus f on the
16
same set of axes.
X1 ( f ) = 2[δ ( f − 2) + δ ( f + 2)]
where Ts =
1
1 



X 2 ( F ) = 2 comb F −  + comb F +  


8
8 

∞
  f 1 
 f 1 
X 2 (Ts f ) = 2 ∑ δ  − − k + δ  + − k 
 16 8 
 16 8  
k =−∞ 
∞
Ts X 2 (Ts f ) = 2 ∑ [δ ( f − 2 − 16 k ) + δ ( f + 2 − 16 k )]
k =−∞
5-22
M. J. Roberts - 8/16/04
|X ( f )|
|T X (T f )|
1
s 2
2
s
2
...
-16
16
f
...
-16
Phase of X1( f )
16
f
Phase of TsX2(Tsf )
π
π
...
-16
16
f
-16
-π
16
...
f
-π
40. Sketch the magnitude and phase of the DTFT of
 n
sinc 
 16 
x1[ n ] =
4
and of the DTFS of
 n
sinc 
 16 
x2[n] =
∗ comb 32 [ n ] .
4
For comparison purposes sketch X1 ( F ) versus F and N 0 X 2 [ k ] versus kF0 on the same set
of axes.
Similar to Exercise 38.
41. A system is excited by a signal,
 t
x( t) = 4 rect  
 2
y( t) = 4 (1 − e −( t +1) ) u( t + 1) − 4 (1 − e −( t −1) ) u( t − 1)
and its response is
[
]
y( t) = 10 (1 − e −( t +1) ) u( t + 1) − (1 − e −( t −1) ) u( t − 1) .
What is its impulse response?
5-23
M. J. Roberts - 8/16/04
Find the CTFT of both x and y. Take their ratio which is the transfer function, H.
Find the inverse transform of H which is h, the impulse response.
5
h( t) = e − t u( t)
2
42. Sketch the magnitudes and phases of the CTFT’s of the following functions.
(a)
g( t) = 5δ ( 4 t)
(c)
g( t) = u(2 t) + u( t − 1)
(b)
 t − 3
 t + 1
g( t) = comb

 − comb
 4 
 4 
Since u ( 2t ) has the same value as u ( t ) for any t, u ( 2t ) = u ( t ) and their
transforms must also be equal when using the time scaling property.
(d)
g( t) = sgn( t) − sgn(− t)
(e)
 t − 1
 t + 1
g( t) = rect 

 + rect 
 2 
 2 
 t − 1 F
 t + 1
←→ 2 sinc ( 2 f ) e j 2 π f + 2 sinc ( 2 f ) e− j 2 π f
+ rect 
rect 


 2 
 2 
 t − 1 F
 t + 1
←→ 4 sinc ( 2 f ) cos ( 2π f )
+ rect 
rect 

 2 
 2 
Using the definition of the sinc function,
sin ( 2π f ) cos ( 2π f )
 t − 1 F
 t + 1
←→ 4
+ rect 
rect 


 2 
 2 
2π f
Using sin ( x ) cos ( y ) =
1
sin ( x − y ) + sin ( x + y ) 
2
1
sin ( 0 ) + sin ( 4π f )  sin ( 4π f )
 t − 1 F
 t + 1
2
←

→
rect 
+
rect
= 4 sinc ( 4 f )
=
4


 2 
2 
πf
2π f
5-24
M. J. Roberts - 8/16/04
(f)
 t
g( t) = rect  
 4
 t
Same answer as in (e) because the function, g( t) = rect   is the same as
 4
the
 t − 1
 t + 1
function, g( t) = rect 
 .
 + rect 
 2 
 2 
(g)
 t
 t
g( t) = 5 tri  − 2 tri 
 5
 2
3
 t
 t
g( t) = rect   ∗ rect  
 8
 2
2
(h)
43. Sketch the magnitudes and phases of the CTFT’s of the following functions.
(b) rect(4 t) ∗ 4δ ( t)
(a) rect ( 4t)
(c) rect(4 t) ∗ 4δ ( t − 2)
(d) rect ( 4 t) ∗ 4δ (2 t)
1
1
 f
 f
F
rect ( 4 t ) ∗ 4δ ( 2t ) ←
→ sinc   × 2 = sinc  
 4
 4
4
2
F
rect ( 4 t ) ∗ 4δ ( 2t ) ←
→=
1
ω 
sinc  
 8π 
2
|
( 4f ) |
1
sinc
2
1
2
-4
4
f
( 4f )
1
sinc
2
rect(4t)∗4δ(2t)
π
2
-1
8
1
8
t
-4
4
f
−π
(e) rect ( 4t) ∗ comb( t)
1
1
 f ∞
 f
F
rect ( 4 t ) ∗ comb ( t ) ←
→ sinc   comb ( f ) = sinc   ∑ δ ( f − k )
 4  k = −∞
 4
4
4
5-25
M. J. Roberts - 8/16/04
F
rect ( 4 t ) ∗ comb ( t ) ←
→
F
rect ( 4 t ) ∗ comb ( t ) ←
→
1 ∞
 k
sinc   δ ( f − k )
∑
 4
4 k = −∞
1 ∞
 π ∞
 k
 k  ω
sinc   δ 
− k  = ∑ sinc   δ (ω − 2π k )
∑
 2 k = −∞
 4
 4   2π
4 k = −∞
f
| 41 sinc( 4 ) comb( f )|
1
4
-4
f
4
f
1
4 sinc 4
( ) comb( f )
rect(4t)∗comb(t)
1
...
-2
-1
-1 1
88
π
...
-4
t
1
4
−π
(f) rect ( 4 t) ∗ comb( t − 1)
Same as part (e).
(g) rect ( 4 t) ∗ comb(2 t)
(h) rect ( t) ∗ comb(2 t)
CTFT is δ ( f ) .
44. Plot these signals over two periods centered at t = 0 .
(a)
x( t) = 2 cos(20πt) + 4 sin(10πt) + 3 cos(−20πt) − 3 sin(−10πt)
(b)
x( t) = 5 cos(20πt) + 7 sin(10πt)
Compare the results of parts (a) and (b).
5-26
f
M. J. Roberts - 8/16/04
This Exercise is intended to show the equivalence between positive and
negative frequencies.
45. A periodic signal has a period of four seconds.
(a)
zero?
What is the lowest positive frequency at which its CTFT could be non-
(b)
What is the next-lowest positive frequency at which its CTFT could be
nonzero?
46. Sketch the magnitude and phase of the CTFT of each of the following signals (ω
form):
x(t)
0.1
−20
(a)
(b)
20
t
 t + 5
x( t) = 3 rect 

 10 
| 30sinc( 5ωπ ) e |
j5ω
30
2π
10
2π
10
ω
j5ω
30sinc( 5ω
π )e
2π
π
x(t)
3
t
−10
(c)
x( t) =
7
 t
comb 
 5
5
5-27
ω
M. J. Roberts - 8/16/04
|X(jω)|
14π
5
...
...
4π 2π
5 5
x (t ) =
(d)
...
7
...
−10 −5
ω
X(jω)
x(t)
...
2π 4π
5 5
4π 2π
5 5
t
5 10
...
2π 4π
5 5
ω
7
 t − 2
comb 
 5 
5
Compare this CTFT with the CTFT of
7
 t + 3
comb 
. Since the two time 5 
5
domain
signals are the same, the two CTFT’s must be the same also. Are they?
47. Sketch the inverse CTFT’s of the following functions:
|X( f )|
20
|X( f )|
20
-4
-4
f
4
f
4
-4
(b)
(a)
| X( f ) |
2
−5
5
f
X( f )
π
−5
-π
f
X( f )
π
2
X( f )
-4
4
5
f
(c)
5-28
π
2
4
f
M. J. Roberts - 8/16/04
(d)
X( f ) = 8δ ( f ) + 5δ ( f − 5) + 5δ ( f + 5)
| X( f ) |
8
5
8+10cos(10πt)
−5
f
5
18
X( f )
−5
2
5
f
5
-2
1
5
1
5
2
5
t
48. Find the inverse CTFT of this real, frequency-domain function (Figure E48) and
sketch it. (Let A = 1, f1 = 95 kHz and f 2 = 105 kHz .)
X( f )
Α
-f2
-f 1
f
f1 f2
Figure E48 A real frequency-domain function
x(t)
x 10
4
2
1
0
-5
-4
-3
-2
-1
0
1
2
3
4
-1
5
-4
x 10
t
-2
49. Find the CTFT (either form) of this signal (Figure E49) and sketch its magnitude and
phase versus frequency on separate graphs. (Let A = − B = 1 and let t1 = 1 and t2 = 2 .)
Hint: Express this signal as the sum of two functions and use the linearity property.
x(t)
A
- t2 - t1
t1 t2
B
5-29
t
M. J. Roberts - 8/16/04
Figure E49 A CT function
|X( f )|
3
-4
4
f
Phase of X( f )
π
-4
4
f
-π
50. In many communication systems a device called a “mixer” is used. In its simplest
form a mixer is simply an analog multiplier. That is, its response signal, y(t), is the
product of its two excitation signals. If the two excitation signals are
x1 ( t) = 10 sinc(20 t)
x 2 ( t) = 5 cos(2000πt)
and
plot the magnitude of the CTFT of y(t), Y( f ) , and compare it to the magnitude of the
CTFT of x1 ( t) . In simple terms what does a mixer do?
|X1( f )|
|Y( f )|
5
4
1
2
f
-10 10
-1010 -990
990 1010
51. Sketch a graph of the convolution of the two functions in each case:
rect(t) * rect(t)
1
(a) rect ( t) ∗ rect ( t)
−1
1
5-30
t
f
M. J. Roberts - 8/16/04
rect(t- 12 ) * rect(t+ 12 )
1
−1
 1
 1
(b) rect  t −  ∗ rect  t + 
 2
 2
(c)
t
1
tri( t) ∗ tri( t − 1)
This is a very challenging problem. It cannot be done using the transforms and tables in
Appendix E but must be done in the time domain.
1
tri(t)
-1
tri(t-1)
1
1
tri(t-τ)
t-1 -1 t
t+1 1
2
t
tri(τ-1)
2
τ
For t < -1, the non-zero portions of the two functions do not overlap and the convolution is zero.
For t > 3, the non-zero portions of the two functions do not overlap and the convolution is zero.
For -1 < t < 0:
The non-zero portions overlap for 0 < τ < t+1 and, in that range of τ,
tri( t − τ ) = t + 1 − τ and tri(τ − 1) = τ
Therefore, for -1 < t < 0,
5-31
M. J. Roberts - 8/16/04
tri( t) ∗ tri( t − 1) =
t +1
t +1
∫ (t + 1 − τ )τdτ = ∫ [(t + 1)τ − τ ]dτ
2
0
0
t +1

τ2 τ3 
(t + 1) (t + 1) (t + 1)
tri( t) ∗ tri( t − 1) = ( t + 1) −  =
−
=
2
3 0
2
3
6

3
3
3
For 0 < t < 1:
tri(t-τ) 1
tri(τ-1)
τ
t 1 t+1 2
-1 t-1
tri( t) ∗ tri( t − 1) =
∞
∫ tri(t − τ ) tri(τ )dτ
−∞
The non-zero portions overlap for 0 < τ < t+2 and, in that range of τ, there are three cases to
consider, 0 < τ < t, t < τ < 1 and 1 < τ < t+1. Therefore
t
1
t +1
0
t
1
tri( t) ∗ tri( t − 1) = ∫ tri( t − τ ) tri(τ ) dτ + ∫ tri( t − τ ) tri(τ ) dτ +
∫ tri(t − τ ) tri(τ )dτ
Case 1:0 < τ < t
tri( t − τ ) = 1 − t + τ and tri(τ − 1) = τ
Case 2:t < τ < 1
tri( t − τ ) = 1 + t − τ and tri(τ − 1) = τ
Case 3:1 < τ < t+1
tri( t − τ ) = 1 + t − τ and tri(τ − 1) = 2 − τ
Therefore
t
1
t +1
0
t
1
tri( t) ∗ tri( t − 1) = ∫ (1 − t + τ )τdτ + ∫ (1 + t − τ )τdτ +
t
[
]
1
[
]
tri( t) ∗ tri( t − 1) = ∫ (1 − t)τ + τ dτ + ∫ (1 + t)τ − τ dτ +
0
2
t
5-32
2
∫ (1 + t − τ )(2 − τ )dτ
t +1
∫ [2(1 + t) − 2τ − (1 + t)τ + τ ]dτ
2
1
M. J. Roberts - 8/16/04
t
t +1
1

τ2 τ3  
τ2 τ3  
τ2 τ3 
tri( t) ∗ tri( t − 1) = (1 − t) +  + (1 + t) −  + 2(1 + t)τ − τ 2 − (1 + t) + 
2
3 0 
2
3 t 
2
3 1


t2 t3  
t2 t3 
1 1
tri( t) ∗ tri( t − 1) = (1 − t) +  + (1 + t) − − (1 + t) + 
2 3
2 3
2 3 


1 1
(t + 1) 2 (t + 1) 3
2
2
+
− 2(1 + t) + 1 + (1 + t) − 
+ 2(1 + t) − ( t + 1) − (1 + t)
2 3
2
3

 t2  1
t 2 2t 3
(t + 1)
2
tri( t) ∗ tri( t − 1) = (1 − t) +
− (1 + t)1 +  + + (1 + t) −
6
2
3
2 3

tri( t) ∗ tri( t − 1) = −
3
t3 t2 t 1
+ + +
2 2 2 6
For the remaining regions of t, the convolution simply repeats with even symmetry about the
point, t = 1. The analytical solutions can be found by the following successive changes of
variable:
t → t + 1 , t → −t , t → t − 1
These three successive changes of variable can be condensed into one,
t → −t + 2
Then, for 1 < t < 2,
 (2 − t) 3 (2 − t) 2 (2 − t) 1 
 t3 t2 t 1 
tri( t) ∗ tri( t − 1) = − + + + 
= −
+
+
+ 
2
2
2
6
 2 2 2 6  t →− t + 2 
and, for 2 < t < 3,
 ( t + 1) 3 
[(− t + 2) + 1]
tri( t) ∗ tri( t − 1) = 
=

6
 6  t →− t + 2
5-33
3
M. J. Roberts - 8/16/04
tri(t) * tri(t -1)
0.8
-5
5
t
(d) 3δ ( t) ∗ 10 cos( t)
(e) 10comb( t) ∗ rect ( t)
(f) 5comb( t) ∗ tri( t)
52. In electronics, one of the first circuits studied is the rectifier. There are two forms, the
half-wave rectifier and the full-wave rectifier. The half-wave rectifier cuts off half of
an excitation sinusoid and leaves the other half intact. The full-wave rectifier reverses
the polarity of half of the excitation sinusoid and leaves the other half intact. Let the
excitation sinusoid be a typical household voltage, 120 Vrms at 60 Hz, and let both
types of rectifiers alter the negative half of the sinusoid while leaving the positive half
unchanged. Find and plot the magnitudes of the CTFT’s of the responses of both
types of rectifiers (either form).
Half-Wave Case:
Full-Wave Case:
x( t) = 120 2 cos(120πt)[rect (120 t) ∗ 60 comb(60 t)]
x( t) = 120 2 cos(120πt)[2 rect (120 t) ∗ 60 comb(60 t) − 1]
|X( f )|
|X( f )|
30 2
60 2
f
f
60
Half-Wave
120
Full-Wave
53. Find the DTFT of each of these signals:
5-34
M. J. Roberts - 8/16/04
n
(a)
 1
x[ n ] =   u[ n − 1]
 3
X( jΩ) =
∞
∑ x[n]e
− jΩn
n =−∞
∞
∞
 1
 1
= ∑   u[ n − 1]e − jΩn = ∑   e − jΩn
 
 
n =−∞ 3
n =1 3
∞
 1
X( jΩ) = ∑  
 
m =0 3
e − jΩ
X( jΩ) =
3
n
m +1
e
n
− jΩ( m +1)
m
 e − jΩ 
= ∑

m =0  3 
 e − jΩ 
e − jΩ
∑  = 3
m =0  3 
∞
∞
m +1
e − jΩ
1
=
e − jΩ 3 − e − jΩ
1−
3
Alternate Solution:
n
n
 1
 1
x[ n ] =   u[ n − 1] =   u[ n ] − δ [ n ]
 3
 3
Using
F
α n u [ n ] ←
→
1
1 − α e− jΩ
and
n
1
 1
x[ n ] =   u[ n ] − δ [ n ] =
 3
3
F
δ [ n ] ←
→1
1
−1
e − jΩ
1−
3
 e − jΩ 
e − jΩ
1 − 1 −

3 

e − jΩ
3
x[ n ] =
=
=
e − jΩ
e − jΩ 3 − e − jΩ
1−
1−
3
3
Second Alternate Solution:
1  1
x[ n ] =  
3  3
X( jΩ) =
n −1
e − jΩ
1
1
e − jΩ =
3 − e − jΩ
3 1 − 1 e − jΩ
3
 π   1
x[ n ] = sin n   u[ n − 2]
 4   4
n
(b)
u[ n − 1]
5-35
M. J. Roberts - 8/16/04
 2π ( n − 2) 
 πn 
 2πn 
sin  = sin
 = cos

 4
 8 


8
 2π ( n − 2)   1 
 1
x[ n ] =   cos
  
 4

 4
8
2
F
α n cos ( Ω0 n ) u [ n ] ←
→
n −2
u[ n − 2]
1 − α cos ( Ω0 ) e− jΩ
, α <1
1 − 2α cos ( Ω0 ) e− jΩ + α 2 e− j 2 Ω
1
π 
2 − jΩ
cos  e − jΩ
1−
e
 4
e − j2Ω
4
 1  − j2Ω
8
X( jΩ) =   e
=
 4
1 π 
1
16
2 − jΩ 1 − j 2Ω
1 − cos  e − jΩ + e − j 2Ω
1
−
e + e
2  4
16
4
16
1−
2
Alternate Solution:
x[ n ] =
e
j
π
n
4
−e
j2
−j
π
n
4
 j π  n  − j π  n
1  e 4  e 4  
 1
−
u[ n − 2]
  u[ n − 2] =  
 4
j 2   4   4  
 
 

n
  j π  2  j π  n − 2  − j π  2  − j π  n − 2
1 e 4  e 4 
e 4  e 4  
−
x[ n ] =  
u[ n − 2]
 4   4  
j2   4   4 
 
 


 

n
 j π4 
F
 e  u [ n ] ←
→
 4 


 j π4 
e 
 4 


1
j
π
e 4 − jΩ
1−
e
4
n−2
e− j 2 Ω
F
u [ n − 2 ] ←
→
j
1−
π
4
e
e− jΩ
4


2
π 2
−
j



  j π4 
− j 2Ω
4
1  e 
e − j 2Ω
e
e


−
X( jΩ) =
π
π


−
j
j
 4 
j2  4 
e 4 − jΩ 
e 4 − jΩ 


 
1−
1−
e
e 


4
4
5-36
M. J. Roberts - 8/16/04
X( jΩ) =
e − j 2Ω
j2
 j π4 
e 
 4 


2
π
−j
  − j π4 

4
e 
1 − e
e − jΩ  − 
  4 

4

 

π
π
j
−j


4
4
e
e
1 −
e − jΩ  1 −


4
4


2
−j
2
π
2
π
j


4
e
−
j
Ω
1 −
e 


4



− jΩ 
e


2
2
j
π
 j π4   j π4  e 4 − jΩ  − j π4   − j π4  e 4 − jΩ
e − e  + e 
e
e  − e 

 
 4
e − j 2Ω     4
X( jΩ) =
π
−j 
j 32
 j π4
− j 2Ω
4
e
e
 e − jΩ + e
+
1− 
 4
4 
16


e − j 2Ω e
X( jΩ) =
j 32
π
j
2
−e
π
−j
2
−
e
j
π
4
e − jΩ +
e
−j
π
4
4
4
1  π  − jΩ e − j 2Ω
1 − cos  e +
2  4
16
e − jΩ
π
−j 
 j π4
4
e
e
−
 e − jΩ
j2 − 


4


e − j 2Ω
X( jΩ) =
e − j 2Ω
1 π 
j 32
1 − cos  e − jΩ +
2  4
16
e − j 2Ω
X( jΩ) =
16
1  π  − jΩ
2 − jΩ
sin  e
1−
e
− j 2Ω


e
4
4
8
=
1  π  − jΩ e − j 2Ω
16
2 − jΩ e − j 2Ω
1 − cos  e +
1−
e +
2  4
16
4
16
1−
(c)
 2π ( n − 4 ) 
 2πn 
x[ n ] = sinc
 ∗ sinc

 8 


8
(d)
 2πn 
x[ n ] = sinc 2 

 8 
Using
5-37
M. J. Roberts - 8/16/04
 n F
sinc   ←
→ wrect ( wF ) ∗ comb ( F )
 w
 n
Find the DTFT of sinc   . Then periodically convolve it with itself.
 w
X(F ) =
8  8 
tri 
F  ∗ comb ( F )
2π  2π 
54. Sketch the magnitudes and phases of the DTFT’s of the following functions:
(b) rect 2 [ n ] ∗ (−5δ [ n ])
(a) rect 2 [ n ]
(c) rect 2 [ n ] ∗ 3δ [ n + 3]
(d) rect 2 [ n ] ∗ (−5δ [ 4 n ]) = rect 2 [ n ] ∗ (−5δ [ n ])
Remember, there is no scaling property for the discrete-time impulse.
(e) rect 2 [ n ] ∗ comb 8 [ n ]
Since this function is periodic, its DTFT must contain only impulses.
|X( F )|
0.625
-1
1
F
Phase of X( F )
π
-1
1
-π
(f) rect 2 [ n ] ∗ comb 8 [ n − 3]
Similar to (e).
5-38
F
M. J. Roberts - 8/16/04
(g) rect 2 [ n ] ∗ comb 8 [2 n ] = rect 2 [ n ] ∗
rect 2 [ n ] ∗ comb 8 [2 n ] = rect 2 [ n ] ∗
∞
∞
∑ δ[2n − 8m] = rect [n] ∗ ∑ δ[2(n − 4 m)]
2
m =−∞
m =−∞
∞
∑ δ[n − 4 m] = rect [n] ∗ comb [n]
m =−∞
2
4
(h) rect 2 [ n ] ∗ comb 5 [ n ]
|X( F )|
1
-1
1
F
Phase of X( F )
π
-1
1
F
-π
55. Sketch the inverse DTFT’s of these functions.
1

(a) X( F ) = comb( F ) − comb F − 

2
F
Using 1←
→ comb ( F ) and e j 2 π F0 n x [ n ] ←
→ X ( F − F0 )
1

F
1 − e jπ n ←
→ comb ( F ) − comb  F − 

2
e
j
πn
2
πn
+j
 F
 − j π2n

2
e
−
e
 ←→ comb ( F ) − comb  F −

− j 2e
−2 e
j
j
πn
2
1

2
1
 πn F

sin   ←
→ comb ( F ) − comb  F − 
 2 

2
π
( n +1)
2
1

 πn F
sin   ←
→ comb ( F ) − comb  F − 

 2 
2
5-39
M. J. Roberts - 8/16/04
1
 π

π
   πn F

→ comb ( F ) − comb  F − 
−2  cos  ( n + 1) + j sin  ( n + 1)  sin   ←

2
  2 

2
 2




 π
1

  πn
π
  πn  F
−2  cos  ( n + 1) sin   + j sin  ( n + 1) sin    ←→ comb ( F ) − comb  F − 

2
  2 
2
  2 
2


=0
 πn


− sin  
 2 


1

 πn F
2 sin 2   ←
→ comb ( F ) − comb  F − 

 2 
2
x[n]
2
-12
12
n
1
1


(b) X( F ) = j comb F +  − j comb F − 


8
8

 
(c) X( F ) = sinc10 F −
 

1 
1
 
  + sinc10 F +    ∗ comb( F )
 
4 
4
Use
T
  πt   T
 t



sinc  ∗ f 0 comb( f 0 t) = wf 0 cos  +  0 − 1 drcl f 0 t, 0 − 1 
 w

w 
  w  w 
from Appendix A (because
X( F ) =
T0
is an integer).
2w
1 
1
1 
1

1 




cos10π  F −   + 9 drcl F − , 9 + cos10π  F +   + 9 drcl F + , 9 

4 
4 
4
4 
10  
∫ cos (10π F ) e
1
j 2 π Fn
F
dF ←
→ cos (10π F )
5-40
M. J. Roberts - 8/16/04
(
)
1
F
e j10 π F + e− j10 π F e j 2 π Fn dF ←
→ cos (10π F )
∫
1
2
1  j 2π F (n + 5)
F
→ cos (10π F )
e
dF + ∫ e j 2 π F ( n − 5 )dF  ←

1
2  ∫1


1  ∫1  cos ( 2π F ( n + 5 )) + j sin ( 2π F ( n + 5 ))  dF  F
←→ cos (10π F )
2  +  cos ( 2π F ( n − 5 )) + j sin ( 2π F ( n − 5 ))  dF 
 
 ∫1 
These integrals are zero unless n = ±5 . Therefore
1
F
δ [ n + 5 ] + δ [ n − 5 ]) ←
→ cos (10π F ) .
(
2
F
Then using e j 2 π F0 n x [ n ] ←
→ X ( F − F0 ) ,
F
rect 4 [ n ] ←
→ 9 drcl ( F, 9 )
Combining inverse transforms,
1
F
δ [ n + 5 ] + δ [ n − 5 ]) + rect 4 [ n ] ←
→ cos (10π F ) + 9 drcl ( F, 9 ) .
(
2
F
Then, using e j 2 π F0 n x [ n ] ←
→ X ( F − F0 )
e
j
πn
2
1
1 

1


 F
 (δ [ n + 5 ] + δ [ n − 5 ]) + rect 4 [ n ] ←→ cos  10π  F −   + 9 drcl  F − , 9 



4
4 
2

and
e
−j
πn
2
1
1 

1


 F
 (δ [ n + 5 ] + δ [ n − 5 ]) + rect 4 [ n ] ←→ cos  10π  F +   + 9 drcl  F + , 9  .



4
4 
2

Then, finally
5-41
M. J. Roberts - 8/16/04
 
1
1  
 j π2n  1


 
−
+
π
F
F
−
cos
10
9
drcl
, 9 
e
δ
n
+
+
δ
n
−
+
rect
n
5
5
[
]
[
]
[
]
(
)








4



4
4  
2
1
1  


F
 ←→ 


10  − j π2n  1
10 
1 
1




+ cos  10π  F +   + 9 drcl  F + , 9 
 (δ [ n + 5 ] + δ [ n − 5 ]) + rect 4 [ n ]
 + e




4  
4
2

The impulses on the left side cancel and we get
 
1
1  


cos  10π  F −   + 9 drcl  F − , 9  
4
4
rect 4 [ n ]

 πn F 1 
→ 
cos   ←

 2 
5
10 
1
1 



+ cos  10π  F +   + 9 drcl  F + , 9 




4
4  
x[n]
0.2
-12
12
n
-0.2
 
(d) X( F ) = δ  F −
 
1
3
5 


 + δ  F −  + δ  F −   ∗ comb(2 F )




4
16
16  
Express comb ( 2F ) as
1

comb ( F ) + comb  F −


2
1 
 . Then do the convolution.
2  
Then do the inverse transform and simplify.


1
3
5 




comb F −  + comb F −  + comb F −  
4
16
16 
1 

X( F ) = 

2 
1 1
3 1
5 1  



+ comb F − −  + comb F − −  + comb F − −  
 



4 2
16 2 
16 2  
 πn
 3π n 
cos   + cos 
 2 
 8 
1
3
5 




comb  F −  + comb  F −  + comb  F −  




4
16 
16  
 5π n  F 1 
+ cos 
←→ 


 8 
2
3
11 
13  



+ comb  F −  + comb  F −  + comb  F − 




4
16 
16  
5-42
M. J. Roberts - 8/16/04
56. Using the relationship between the CTFT of a signal and the CTFS of a periodic
extension of that signal, find the CTFS of
 t
 t 1
x( t) = rect   ∗ comb 
 w  T0
 T0 
and compare it with the table entry.
57. Using the relationship between the DTFT of a signal and the DTFS of a periodic
extension of that signal, find the DTFS of
rect N w [ n ] ∗ comb N 0 [ n ]
and compare it with the table entry.
5-43