Solution to ECE 315 Final Examination
Su05
1.
For each active filter below find the numerical value of the magnitude of the transfer
function H ( f ) = Vo ( f ) / Vi ( f ) at the two extremes, f = 0 and f → +∞ . (All resistors are
1 ohm and all capacitors are 1 F.)
+
vi (t)
-
(a)
+
vo (t)
vx (t)
ii (t)
-
R f if (t)
Ri
H ( 0 ) = 1 or 0 dB
H ( +∞ ) = 2
H ( 0 ) = 0 or −∞ dB
H ( +∞ ) = −1 = 1 or 0 dB
Ci
Rf
(b)
i i(t) R i
Ci
+
if (t)
vx (t)
+
vi (t)
vo (t)
-
-
Rf
H ( 0 ) = −1 = 1 or 0 dB
C
(c)
+
i i(t) R i vx (t)
H ( +∞ ) = 0 or −∞ dB
if (t)
+
vi (t)
vo (t)
-
-
Rf
(d)
+
i i(t) R i vx (t)
+
vi (t)
vo (t)
-
-
H ( 0 ) = −1 = 1 or 0 dB
-
Ci
H ( +∞ ) = −∞ = ∞ or +∞ dB
+
vi (t)
+
if (t) v-o (t)
vx (t)
ii (t)
(e)
if (t)
Ri
Rf
Cf
H ( 0 ) = 2 or 6 dB
H ( +∞ ) = 1 or 0 dB
2.
In the circuit below R = 100 Ω , C = 5 µ F , L = 10 m H .
+
R
vin(t)
+
C
L vout(t)
(a)
Find the frequency at which the magnitude of the transfer function is a maximum.
H ( jω ) =
Vo ( jω )
( jω L / jω C ) / ( jω L + 1 / jω C ) =
( jω L / jω C )
=
Vi ( jω ) ( jω L / jω C ) / ( jω L + 1 / jω C ) + R ( jω L / jω C ) + R ( jω L + 1 / jω C )
H ( jω ) =
( jω )
jω L
2
RLC + jω L + R
=
jω
1
2
RC ( jω ) + jω / RC + 1 / LC
Magnitude is a maximum at resonance which is at ω =
f = 711.8 .
(b)
-
1
=
LC
1
50 × 10 −9
= 4472 or
Find the non-zero frequency at which the phase of the transfer function is zero.
Same answer, ω = 4472 or f = 711.8 .
(c)
Find the phase shift of the transfer function just above and just below f = 0 .
jω
L which is π / 2 radians.
R
jω
Just below f = 0 the phase is the phase of H ( jω ) ≅
L which is −π / 2 radians.
R
Just above f = 0 the phase is the phase of H ( jω ) ≅
(d)
Classify this filter as a practical approximation to the ideal lowpass, highpass,
bandpass or bandstop filter.
Bandpass
3.
In the system below let xt ( t ) = 3 sin (1000π t ) , let fc = 5000 and let the lowpass
filter (LPF) be ideal with a transfer function magnitude of one in its passband.
yt (t) = x r(t)
x t(t)
cos(2πfct)
yd (t)
LPF
yf (t)
cos(2πfct)
(a)
Using the principle that the signal power of a sum of sinusoids plus a constant is the
sum of the signal power of the constant and the signal powers in the individual sinusoids if
their frequencies are all different, find the signal power of y t ( t ) . (The signal power of a
single sinusoid is one-half of the square of its amplitude.)
y t ( t ) = 3 sin (1000π t ) cos (10000π t )
Yt ( f ) =
j3
1
δ ( f + 500 ) − δ ( f − 500 )  ∗ δ ( f − 5000 ) + δ ( f + 5000 ) 
2
2
Yt ( f ) =
j3
δ ( f − 4500 ) + δ ( f + 5500 ) − δ ( f − 5500 ) − δ ( f + 4500 ) 
4 
yt (t ) =
3
 − sin ( 9000π t ) + sin (11000π t ) 
2
Py = 2 × ( 3 / 2 ) / 2 =
2
(b)
9
= 2.25
4
Find the signal power of y d ( t ) .
Yd ( f ) =
j3
δ ( f − 4500 ) + δ ( f + 5500 ) − δ ( f − 5500 ) − δ ( f + 4500 ) 
4 
1
∗ δ ( f − 5000 ) + δ ( f + 5000 ) 
2
Yd ( f ) =
j 3 δ ( f − 9500 ) + δ ( f + 500 ) − δ ( f − 10500 ) − δ ( f − 500 ) 


8  +δ ( f + 500 ) + δ ( f + 10500 ) − δ ( f − 500 ) − δ ( f + 9500 ) 
Yd ( f ) =
j 3 δ ( f − 9500 ) − δ ( f + 9500 ) + 2δ ( f + 500 ) − 2δ ( f − 500 ) 


8  +δ ( f + 10500 ) − δ ( f − 10500 )

yd (t ) =
3
 − sin (19000π t ) + sin ( 21000π t ) + 2 sin (1000π t ) 
4
Py = 2 × ( 3 / 4 ) / 2 + ( 6 / 4 ) / 2 =
2
(c)
kHz.
(d)
Hz.
2
9 18 27
+
=
= 1.6875
16 16 16
Find the signal power of y f ( t ) if the cutoff frequency of the lowpass filter is 1
Yf ( f ) =
j3
 2δ ( f + 500 ) − 2δ ( f − 500 ) 
8 
yf ( f ) =
3
18
sin (1000π t ) ⇒ Py =
= 1.125
2
16
Find the signal power of y f ( t ) if the cutoff frequency of the lowpass filter is 100
Py = 0
4.
In the system below let xt ( t ) = 3 sin (1000π t ) , let m = 1 , let A = 3 , let fc = 5000
and let the lowpass filter (LPF) be ideal with a transfer function magnitude of one in its
passband.
yt (t) = x r(t)
m
x t(t)
A
cos(2πfct)
yd (t)
LPF
yf (t)
cos(2πfct)
(a)
Using the principle that the signal power of a sum of sinusoids plus a constant is the
sum of the signal power of the constant and the signal powers in the individual sinusoids if
their frequencies are all different, find the signal power of y t ( t ) . (The signal power of a
single sinusoid is one-half of the square of its amplitude.)
y t ( t ) = 3( sin (1000π t ) + 1) cos (10000π t )
j3

 1
Yt ( f ) =  3δ ( f ) + δ ( f + 500 ) − δ ( f − 500 )   ∗ δ ( f − 5000 ) + δ ( f + 5000 ) 
2

 2
Yt ( f ) =
3
δ ( f − 5000 ) + δ ( f + 5000 ) 
2
j3
+ δ ( f − 4500 ) + δ ( f + 5500 ) − δ ( f − 5500 ) − δ ( f + 4500 ) 
4
y t ( t ) = 3 cos (10000π t ) +
(
)
3
sin (11000π t ) − sin ( 9000π t ) 
2
Py = 32 / 2 + 2 × ( 3 / 2 ) / 2 =
(b)
2
9 9 27
+ =
= 6.75
2 4 4
Find the signal power of y d ( t ) .

3

 2 δ ( f − 5000 ) + δ ( f + 5000 ) 
Yd ( f ) = 

+ j 3 δ ( f − 4500 ) + δ ( f + 5500 ) − δ ( f − 5500 ) − δ ( f + 4500 )  

 4 

1
∗ δ ( f − 5000 ) + δ ( f + 5000 ) 
2
1
3

 2 δ ( f − 5000 ) + δ ( f + 5000 )  ∗ 2 δ ( f − 5000 ) + δ ( f + 5000 )  


 j3

Yd ( f ) = + δ ( f − 4500 ) + δ ( f + 5500 ) − δ ( f − 55500 ) − δ ( f + 4500 )  
 4

1


∗ δ ( f − 5000 ) + δ ( f + 5000 ) 


2
3

 4 δ ( f − 10000 ) + 2δ ( f ) + δ ( f + 10000 ) 



Yd ( f ) = 
δ ( f − 9500 ) + 2δ ( f + 500 ) + δ ( f + 10500 )  
+ j 3 

 8  −δ ( f − 10500 ) − 2δ ( f − 500 ) − δ ( f + 9500 )  
yd (t ) =
3
3 3
cos ( 20000π t ) + +  2 sin (1000π t ) + sin ( 21000π t ) − sin (19000π t ) 
2
2 4
Py = ( 3 / 2 ) / 2 + ( 3 / 2 ) + ( 3 / 2 ) / 2 + 2 × ( 3 / 4 ) / 2 =
2
2
Py =
(c)
kHz.
2
9 9 9 9
+ + +
8 4 8 16
18 + 36 + 18 + 9 81
=
= 5.0625
16
16
Find the signal power of y f ( t ) if the cutoff frequency of the lowpass filter is 1
y f (t ) =
(d)
Hz.
2
3 3
9 9 27
2
2
+ sin (1000π t ) ⇒ Py = ( 3 / 2 ) + ( 3 / 2 ) / 2 = + =
= 3.375
2 2
4 8 8
Find the signal power of y f ( t ) if the cutoff frequency of the lowpass filter is 100
y f (t ) =
3
9
2
⇒ Py = ( 3 / 2 ) = = 2.25
2
4
5.
What is the fundamental reason that an ideal filter cannot physically exist, even if the
components in it were ideal?
Ideal filters are not causal.
6.
A system is excited by a sinusoid whose signal power is 0.01 and the response is a
sinusoid of the same frequency with a signal power of 4. What is the magnitude of the
transfer function of the system at the frequency of the sinusoid, expressed in decibels (dB)?
Power Ratio
 4 
10 log10 
= 10 log10 ( 400 ) = 26 dB
 0.01 
7.
A system is excited by a sinusoid whose amplitude is 1 µV and the response is a
sinusoid of the same frequency with an amplitude of 5 V. What is the magnitude of the
transfer function of the system at the frequency of the sinusoid, expressed in decibels (dB)?
Signal Ratio
5


20 log10  −6  = 20 log10 5 × 10 6 = 134 dB
 10 
(
)
8.
An ideal DT lowpass filter has a transfer function magnitude of 3 at Ω = 0 and a
radian cutoff frequency of π / 4 . Find the numerical value of the transfer function
magnitude at
(a)
Ω=π /2
0
All DT filters have periodic frequency responses. π / 2 is greater than π / 4 and
less than 2π − π / 4 which is where the lower corner of the next periodic replica occurs.
(b)
Ω = 7π / 8
0
7π / 8 is greater than π / 4 and less than 2π − π / 4 which is where the lower
corner of the next periodic replica occurs.
(c)
Ω = 22π
3
If the filter’s transfer function is 3 for radian frequencies −π / 4 < Ω < π / 4 it is
also 3 for radian frequencies 2 nπ − π / 4 < Ω < 2 nπ + π / 4 , n an integer.