Solution to ECE Test #7 S07 #1
ω
6
[s]
|H(jω)|
4
1.2
2
-20
2
ω
20
ω
Phase ofH(jω)
π
4
-20
6
6
20
σ
0
4
2
0
2
4
6
-π
This is an “all pass” filter. As moves from zero to infinity each zero vector is exactly
the same length as its corresponding pole vector so the magnitude of the transfer function
never changes.
ω
6
[s]
|H(jω)|
4
25
2
-20
2
ω
20
ω
Phase ofH(jω)
π
4
-20
6
6
20
σ
0
4
2
0
2
4
6
-π
The response must be zero at = 0 . There is one more finite zero than finite poles so at
high frequencies the magnitude must rise linearly with .
ω
6
[s]
|H(jω)|
4
0.7
2
-20
2
ω
20
ω
Phase ofH(jω)
π
4
-20
6
6
20
σ
0
4
2
0
2
4
6
-π
Finite response at = 0 . One more finite pole than finite zeros so the magnitude falls as
1 / at high frequencies and phase approaches / 2 for + .
ω
6
[s]
|H(jω)|
4
1.6
2
-20
2
ω
20
ω
Phase ofH(jω)
π
4
-20
6
6
20
σ
0
4
2
0
2
4
6
-π
Two zeros on the axis. Therefore the magnitude must go to zero at a non-zero finite
value of . Also, the phase is discontinuous at that point. One more finite pole than finite
zeros so the phase must approach / 2 for + .
ω
6
[s]
|H(jω)|
4
2
2
-20
2
ω
20
ω
Phase ofH(jω)
π
4
6
6
20
σ
0
-20
4
2
0
2
4
6
-π
Similar to the last one except the zeros are near but not on the axis. Similar magnitude
response except the magnitude does not go to zero at a non-zero finite value of . One
more finite pole than finite zeros so the phase must approach / 2 for + .
ω
6
[s]
|H(jω)|
4
0.07
2
-20
2
ω
20
ω
Phase ofH(jω)
π
4
-20
6
6
20
σ
0
4
2
0
2
4
6
-π
Four finite poles and no finite zeros. All four poles are near the axis. Therefore there
are four recognizable peaks, one per pole, in the magnitude. Phase goes through and
jumps to + (because of the way MATLAB plots phase) and then goes negative from
there another radians, because of four finite poles and no finite zeros.
ω
6
[s]
|H(jω)|
4
16
2
-20
2
ω
20
ω
Phase ofH(jω)
π
4
-20
6
6
20
σ
0
4
2
0
2
4
6
-π
Two finite poles near the axis so there are two sharp peaks in the magnitude frequency
response. Same number of finite poles and finite zeros so the phase approaches zero at
high frequencies.
ω
6
[s]
|H(jω)|
4
1
2
-20
2
ω
20
ω
Phase ofH(jω)
π
4
-20
6
6
20
σ
0
4
2
0
2
4
6
-π
Two finite zeros near the axis so there are two dips in the magnitude frequency
response. Same number of finite poles and finite zeros so the phase approaches zero at
high frequencies.