Solution to EE 503 Test #3 S03 4/4/03
 t
The CTFT of the CT signal, x( t) = tri  ∗ comb(2 t) , can be expressed as a single
 4
impulse of the form, X( f ) = Aδ ( f ) . What is the numerical value of A?
1.
1
 f
 f
X( f ) = 4 sinc 2 ( 4 f ) × comb  = 2 sinc 2 ( 4 f ) comb 
 2
 2
2
∞
∞
f

X( f ) = 2 sinc 2 ( 4 f ) ∑ δ  − k = 4 ∑ sinc 2 ( 4 f )δ ( f − 2 k )
2

k =−∞
k =−∞
All the impulses fall on nulls of the sinc 2 function except the one at f = 0
and that one has a strength of 4. Therefore A = 4 .
What is the maximum numerical value of the CT function, x( t) , which is the
 f
inverse CTFT of X( f ) = rect   comb( f ) ? ( x( t) can be expressed as a constant plus
 5
some sinusoids.)
2.
 f
X( f ) = rect   comb( f ) = δ ( f + 2) + δ ( f + 1) + δ ( f ) + δ ( f − 1) + δ ( f − 2)
 5
x( t) = 1 + 2 cos(2πt) + 2 cos( 4πt)
Maximum value occurs when the cosines have simulatneous positive peak
values, for example at t = 0. So the maximum value is 5.
3.
 3n 
Find the numerical signal energy of the DT signal, x[ n ] = 25 sinc  .
 11 
Use Parseval’s theorem for DT signals.
X( F ) = 25 ×
11
275
 11 
 11 
rect  F  ∗ comb( F ) =
rect  F  ∗ comb( F )


3 
3
3
3
The total signal energy is the area under the square of the magnitude of X( F ) over exactly
one period. That area is
2
 275  3 226875
X( F ) = 
=
≅ 2291.67 .

 3  11
99
A DT signal, x[ n ] , is formed by sampling a CT signal, x( t) , at a sampling rate of
1
1 



50 Hz. The DTFT of x[ n ] is X( F ) = 10 comb F −  + comb F +   . Find two


4
4 

different CT functions, either of which could be the function, x( t) , that was sampled.
4.
 2πn 
x[ n ] = 20 cos

 4 
The CT signal that was sampled must be of the form, x( t) = 20 cos(2πf 0 t) . When it is

f 
sampled, it is of the form, x[ n ] = 20 cos(2πf 0 nTs ) = 20 cos 2πn 0  . The simplest form of
fs 

f
2πn
, which implies that
x( t) = 20 cos(2πf 0 t) would be found by letting 2πn 0 =
fs
4
f0 1
50
= ⇒ f0 =
= 12.5 and x( t) = 20 cos(25πt) . If the frequency is changed by any
fs 4
4
integer multiple of the sampling rate the samples do not change. Therefore, the most
general form of the CT signal would be
x( t) = 20 cos(2π (12.5 + 50 k ) t)
where k is any integer.
5.
Each practical passive filter below is an approximation to one of the four basic ideal
filter types, lowpass, highpass, bandpass and bandstop. Identify the function of each filter
with one of those four terms.
(a)
Bandstop
(b)
Bandpass
(c)
Highpass
(d)
Lowpass
(e)
Highpass
(f)
Lowpass
(g)
Bandstop
(h)
Bandpass
(a)
(b)
C
+
i i(t)
+
L
vi (t)
vo(t)
R
-
-
+
-
+
C
L vo(t)
-
-
+
+
L
vo(t)
-
+
vi (t)
-
+
vo(t)
R
-
-
+
v i (t)
+
R
vo(t)
-
+
+
+
C
-
+
vo(t)
C
-
(h)
i i(t) R
vi (t)
i i(t) R
-
(g)
i i(t) L
vi (t)
(d)
i i(t) C
(f)
i i(t) R
vi (t)
i i(t) R
vi (t)
(e)
+
(c)
L
vo(t)
-
i i(t) L
vi (t)
-
C
R
+
vo(t)
-
6.
Let the excitation signal, v i ( t) , in the circuit below be of the form,
v i ( t) = cos(2πf 0 t) . Let the component values be Ri = 10 kΩ, R f = 100 kΩ, C = 1 µF and
L = 1 mH. At what numerical value of f 0 will the response signal, v o ( t) , have the largest
possible amplitude and what will that amplitude be? (Assume the operational amplifier is
ideal.)
L
Rf
C
+
i i(t) R i vx (t)
if (t)
+
vi (t)
vo (t)
-
-
The maximum response amplitude occurs at the frequency of maximum gain magnitude.
The gain is
1
H( f ) = −
G f + sC +
Ri
1
j 2πfL
j 2πfL
−(2πf ) LC + j 2πfLG f + 1
1
j 2πf
=−
Ri
RiC − 2πf 2 + j 2πf + 1
( )
R f C LC
2
=−
The maximum gain occurs at resonance where
1
2
− (2πf ) = 0 . Solving for the resonant
LC
1
1
=
= 5.033 kHz .
At resonance the transfer
−3
2π LC 2π 10 × 10 −6
Rf
100
function is H( f ) = −
=−
= −10 and, since the excitation amplitude is one, the
10
Ri
response amplitude is 10.
frequency,
f =
1
In the system below, if x t ( t) = 10, m = , f c = 100 kHz and the lowpass filter is
2
ideal with a cutoff frequency of 1 kHz, what is the numerical value of the response, y f ( t) ?
7.
x t(t)
yd (t)
yt (t) = x r(t)
m
cos(2πfct)
1
yf (t)
LPF
cos(2πfct)
y t ( t) = x r ( t) = 6 cos(2πf c t)


5

y d ( t) = 6 cos (2πf c t) = 3 1 + cos( 4πf c t) = 3 1 + cos( 4 × 10 πt) 
 144244
3
removed by LPF 

y f ( t) = 3
[
2
]
8.
The Nyquist sampling rate is the dividing line between undersampling and
oversampling a signal. For each signal below find its Nyquist rate. If the signal is not
bandlimited just write “infinite”.
(a)
 t
x( t) = −4 sinc 
 8
Nyquist rate =
X( f ) = −32 rect (8 f ) ⇒ f Nyq =
(b)
(c)
 t − 4
x( t) = 25 tri

 2 
1
8
1
8
Nyquist rate = Infinite
X( f ) = 50 sinc 2 (2 f )e − j 8πf
x( t) = 3 cos(100πt) sin(10, 000πt)
Nyquist rate = 10,100
j
3
δ ( f − 50) + δ ( f + 50)] ∗ [δ ( f + 5000) − δ ( f − 5000)]
[
2
2
3
X( f ) = j [δ ( f − 50) + δ ( f + 50)] ∗ [δ ( f + 5000) − δ ( f − 5000)]
4
3
X( f ) = j [δ ( f + 4950) − δ ( f − 5050) + δ ( f + 5050) − δ ( f − 4950)]
4
X( f ) =
(d)
x( t) = −50 sinc(20 t) cos(80πt)
X( f ) = −
Nyquist rate = 100
50
 f  1
rect   ∗ [δ ( f − 40) + δ ( f + 40)]
 20  2
20
5
 f + 40  
 f − 40 
X( f ) = − rect 

 + rect 
 20  


4
20
9.
A signal is sampled 3 times and the samples are {x[0] , x[1] , x[2]} = {1 , − 1 , − 2} .
The discrete Fourier transform (DFT) of this set of samples is X[ k ]. What is the numerical
value of X[1] ?
N −1
X[ k ] = ∑ x[ n ]e
n =0
− j 2π
nk
N
3 −1
⇒ X[1] = ∑ x[ n ]e
− j 2π
n
3
= 1− e
−j
2π
3
− 2e
−j
4π
3
n =0
 1
 1
3 5
3
3
X[ k ] = 1 −  − − j
− 2 − + j
= −j
= 2.5 − j 0.866


2  2
2
2 
 2
 2
10.
A bandlimited periodic CT signal, x( t) , is sampled, at a rate greater than its
Nyquist rate, 4 times over exactly one fundamental period to form the DT signal, x[ n ] .
The discrete Fourier transform (DFT) of that set of samples is
{X[0] , X[1] , X[2] , X[3]} = {2 , 1 − j , 0 , 1 + j} .
(a)
What is the numerical average value of the samples,
{x[0] , x[1] , x[2] , x[3]} ?
The sum of the samples is X[0] = 2 .
1
average of the samples is .
2
(b)
There are four samples.
Therefore the
What is the numerical average signal power of x( t) ?
Since the signal is bandlimited and sampled properly, the CTFS of the signal is
X[ k ] =
1
{(1 + j )δ[k + 1] + 2δ[k ] + (1 − j )δ[k − 1]} .
4
The average signal power is the sum of the squares of these amplitudes which is
2+4+2 1
= .
16
2