Solution of ECE 315 Test 13 F04
1.
An LTI system has a transfer function,
H ( jω ) =
j 3ω − ω 2
.
1000 − 10ω 2 + j 250ω
(a)
Find all the corner frequencies (in radians per second) in a magnitude Bode plot of this transfer function.
Corner frequencies (radians/s): 3, 5, 20 (and 0)
Corner frequencies are the magnitudes of the locations of the non-zero real poles and zeros and the radian resonant
frequencies of any complex poles or zeros.
H ( jω ) =
jω ( jω + 3)
10 ( jω ) + j 250ω + 1000
2
=
1
1
× jω × ( jω + 3) ×
10
( jω + 5 ) ( jω + 20 )
Zeros at jω = 0 and jω = −3 . Poles at jω = −5 and jω = −20 . (There is not actually a “corner” at zero but I accepted
it anyway.)
(b)
At very low and very high frequencies what is the slope of the magnitude Bode plot in dB/decade?
Slope = +20 dB/decade at very low frequencies and 0 dB/decade at very high frequencies
At very low freqencies the transfer function is approximately H ( jω ) = j 3ω / 1000 . This is a frequency-independent gain
times a single differentiator so the slope is +20 dB/decade. At very high frequencies the transfer function is
jω ( jω ) 1
approximately H ( jω ) =
which is a constant so the slope is 0 dB/decade.
2 =
10
10 ( jω )
2.
(a)
Find the transfer function, H ( jω ) =
Vo ( jω )
, of this active filter with Ri = 1000 Ω , Ci = 1µF and
Vi ( jω )
R f = 5000 Ω .
if(t)
i i(t)
Ri
Rf
Ci
+
vx(t)
vi (t)
-
Using the gain formula for an inverting amplifier, H ( jω ) = −
Rf
Ri +
1
jω Ci
+
vo(t)
-
=−
jω R f Ci
jω Ri Ci + 1
=−
j 5 × 10 −3ω
.
j10 −3ω + 1
(b)
Find all the corner frequencies (in radians per second) in a magnitude Bode plot of this transfer function.
Corner frequencies (radians/s): 1000 (and 0)
There is one real pole at j10 −3ω + 1 = 0 ⇒ jω = −1000 and one real zero at jω = 0 . (Again, there is not actually a
“corner” at zero but I accepted it anyway.)
(c)
At very low and very high frequencies what is the slope of the magnitude Bode plot in dB/decade?
Slope = +20 dB/decade at very low frequencies and 0 dB/decade at very high frequencies
At very low freqencies the transfer function is approximately H ( jω ) = − j 5 × 10 −3ω . This is a single differentiator times
a frequency-independent gain of −5 × 10 −3 so the slope is +20 dB/decade. At very high frequencies the transfer function
is approximately H ( jω ) = −5 which is a constant so the slope is 0 dB/decade.
Solution of ECE 315 Test 13 F04
1.
An LTI system has a transfer function,
H ( jω ) =
1200 − 10ω 2 + j 340ω
.
j 6ω − ω 2
(a)
Find all the corner frequencies (in radians per second) in a magnitude Bode plot of this transfer function.
Corner frequencies (radians/s): 6, 4, 30 (and 0)
Corner frequencies are the magnitudes of the locations of the non-zero real poles and zeros and the radian resonant
frequencies of any complex poles or zeros.
H ( jω ) =
10 ( jω ) + j 340ω + 1200
1
1
= 10 × ( jω + 4 ) ( jω + 30 ) ×
×
jω ( jω + 6 )
jω jω + 6
2
Zeros at jω = −4 and jω = −30 . Poles at jω = 0 and jω = −6 . (There is not actually a “corner” at zero but I accepted
it anyway.)
(b)
At very low and very high frequencies what is the slope of the magnitude Bode plot in dB/decade?
Slope = -20 dB/decade at very low frequencies and 0 dB/decade at very high frequencies
At very low freqencies the transfer function is approximately H ( jω ) = 200 / jω . This is a frequency-independent gain
times a single integrator so the slope is -20 dB/decade. At very high frequencies the transfer function is approximately
H ( jω ) = 10 which is a constant so the slope is 0 dB/decade.
2.
(a)
Find the transfer function, H ( jω ) =
Vo ( jω )
, of this active filter with Ri = 2000 Ω , C f = 1µF and
Vi ( jω )
R f = 4000 Ω .
if(t)
i i(t)
+
Rf
Cf
Ri
vx(t)
vi (t)
+
vo(t)
-
Using the gain formula for an inverting amplifier, H ( jω ) = −
Rf +
1
jω C f
Ri
-
=−
jω R f C f + 1
jω Ri C f
=−
j 4 × 10 −3ω + 1
.
j 2 × 10 −3ω
(b)
Find all the corner frequencies (in radians per second) in a magnitude Bode plot of this transfer function.
Corner frequencies (radians/s): 250 (and 0)
There is one real zero at j 4 × 10 −3ω + 1 = 0 ⇒ jω = −250 and one real pole at jω = 0 . (Again, there is not actually a
“corner” at zero but I accepted it anyway.)
(c)
At very low and very high frequencies what is the slope of the magnitude Bode plot in dB/decade?
Slope = -20 dB/decade at very low frequencies and 0 dB/decade at very high frequencies
At very low freqencies the transfer function is approximately H ( jω ) = −500 / jω . This is a single integrator times a
frequency-independent gain of −500 so the slope is -20 dB/decade. At very high frequencies the transfer function is
approximately H ( jω ) = −2 which is a constant so the slope is 0 dB/decade.