Frequency Response
Chapter
10
Chapter 9 dealt with time-domain response of continuous-time
systems. The frequency response of continuous-time systems
gives another view, just as it did for discrete-time systems in
Chapter 6. The concept of frequency response is again motivated
by applying a single sinusoid.
The Frequency Response Function for LTI
Systems
• The output of an LTI system can be given in terms of the convolution integral
y( t) = h(t)*x(t) =
∞
∫–∞ h ( τ )x ( t – τ ) dτ
(10.1)
where we recall that h ( t ) is the (unit) impulse response of a
system
• We choose to start the analysis with a single complex sinusoid
jφ jωt
x ( t ) = Ae e
, –∞ < ω < ∞
(10.2)
• The output is
ECE 2610 Signal and Systems
10–1
The Frequency Response Function for LTI Systems
jφ jωt
y ( t ) = h ( t ) * ( Ae e
∞
=
∫–∞
= ⎛
⎝
)
jφ jω ( t – τ )
h ( τ )Ae e
∞
∫–∞
h ( τ )e
– j ωτ
dτ
jφ jωt
dτ⎞ Ae e
⎠
(10.3)
jφ jωt
= H ( jω )Ae e
• We have thus defined the frequency response of an LTI system as
Frequency Response
H ( jω ) =
Example: h ( t ) = e
– αt
∞
∫–∞
h ( τ )e
– j ωτ
dτ
(10.4)
u(t)
• From the definition
H ( jω ) =
∞
∫–∞
∞
=
∫0
e
e
– ατ
u ( τ )e
– ( jω + α )τ
– ( jω + α )τ
e
= -----------------------– ( jω + α )
∞
0
– j ωτ
dτ
dτ
1 = --------------α + jω
• Given the frequency response we can now plot the magnitude
and phase response just like was done for a discrete-time system
ECE 2610 Signals and Systems
10–2
The Frequency Response Function for LTI Systems
• A major distinction here is that the frequency axis runs from
– ∞ to ∞
• We can use Matlab to do this using either a direct calculation
or the function freqs()
>> help freqs
FREQS Laplace-transform (s-domain) frequency response.
H = FREQS(B,A,W) returns the complex frequency response vector H
of the filter B/A:
nb-1
nb-2
B(s)
b(1)s
+ b(2)s
+ ... + b(nb)
H(s) = ---- = ------------------------------------na-1
na-2
A(s)
a(1)s
+ a(2)s
+ ... + a(na)
given the numerator and denominator
The frequency response is evaluated
vector W (in rad/s). The magnitude
calling FREQS(B,A,W) with no output
coefficients in vectors B and A.
at the points specified in
and phase can be graphed by
arguments.
[H,W] = FREQS(B,A) automatically picks a set of 200 frequencies W on
which the frequency response is computed. FREQS(B,A,N) picks N
frequencies.
• The full understanding of freqs() is not possible until the
Laplace transform is studied in systems and circuits
• The H ( jω ) we have is of the form
H ( jω ) = H ( s )
s = jω
1 = ----------s+α
s = jω
1 = --------------jω + α
• Let α = 2
>>
>>
>>
>>
>>
>>
>>
w = -50:.1:50;
H = freqs([1],[1 2],w);
subplot(211)
plot(w,abs(H))
grid
ylabel('|H(j\omega)|')
subplot(212)
ECE 2610 Signals and Systems
10–3
Response to Real Sinusoid Signals
>>
>>
>>
>>
plot(w,angle(H))
grid
ylabel('\angle H(j\omega)')
xlabel('\omega (rad/s)')
|H(jω)|
0.4
1 H ( jω ) = -------------2 + jω
0.3
0.2
0.1
0
−50
−40
−30
−20
−10
−40
−30
−20
−10
0
10
20
30
40
50
0
10
ω (rad/s)
20
30
40
50
2
∠ H(jω)
1
0
−1
−2
−50
Response to Real Sinusoid Signals
• The frequency response can also be used to find the system
output when the input is a real sinusoid
• Just as in the case of discrete-time systems, when the input is
x ( t ) = A cos ( ω 0 t + φ ), – ∞ < t < ∞
(10.5)
the output is
y ( t ) = A H ( jω 0 ) cos ( ω 0 t + φ + ∠H ( jω 0 ) )
ECE 2610 Signals and Systems
(10.6)
10–4
Ideal Filters
Symmetry of H ( jω )
• Again utilizing what was learned for discrete-time systems,
for a system having real impulse response the following symmetry condition holds
H ( – jω ) = [ H ( jω ) ] * = H * ( jω )
(10.7)
which means that
Magnitude and Phase Symmetry
H ( – j ω ) = H ( jω )
∠H ( – jω ) = – ∠H ( jω )
(10.8)
Response to a Sum of Sinusoids
(10.9)
Periodic Signal Inputs
Ideal Filters
ECE 2610 Signals and Systems
10–5
Simulation of Circuit Implementations
Simulation of Circuit Implementations
• As the study of systems and circuits moves forward in the
courses ECE 2205, 3205, and ultimately electronic circuits,
the need will eventually arise to simulate circuit based realizations of systems
• We now briefly introduce circuit analysis of a simple RC
lowpass filter in terms of frequency response and time
domain simulation for a pulse input and a sinusoid input
• A free cross platform compatible circuit simulator is Qucs,
which stands for quite a universal circuit simulator (http://
qucs.sourceforge.net/)
ECE 2610 Signals and Systems
10–6
Simulation of Circuit Implementations
• AC simulation is the means for obtaining frequency response
type results for a circuit implementation
Input
Output
ac simulation
AC1
Type=log
Start=0.01 Hz
Stop=1
Points=201
V1
U=1 V
R1
R=1 Ohm
C1
C=1 F
Equation
Eqn3
Ampl=abs(Output.v)
Phase=phase(Output.v)
H ( j2πf ) In Terms of Magnitude and Phase
0
1
-20
0.8
Magnitude (blue)
-40
0.6
0.4
-60
Phase (red)
0.2
-80
0
0.1
0.2
0.3
ECE 2610 Signals and Systems
0.4
0.5
0.6
Frequency /(2) (Hz)
0.7
0.8
0.9
1
10–7
Simulation of Circuit Implementations
• Transient simulation is used to obtain pulse input and sinusoidal input time-domain simulations
Input
Output
transient
simulation
V1
U1=0 V
U2=1 V
T1=0
T2=5 s
TR1
Type=lin
Start=0
Stop=15 s
R1
R=1 Ohm
C1
C=1 F
Input = u(t) - u(t - 5)
1
0.8
0.6
Output
0.4
0.2
0
0
2
4
ECE 2610 Signals and Systems
6
8
Time (s)
10
12
14
10–8
Simulation of Circuit Implementations
• Now consider the time domain response to a 1 Hz sinusoidal
input applied starting at t = 0
Output
Input
transient
simulation
V1
U=1 V
R1
R=1 Ohm
C1
C=1 F
TR1
Type=lin
Start=0
Stop=15 s
0.3
Transient
Steady-state
0.2
0.1
0
-0.1
-0.2
0
2
4
ECE 2610 Signals and Systems
6
8
Time (s)
10
12
14
10–9
Simulation of Circuit Implementations
ECE 2610 Signals and Systems
10–10