Frequency Response
Lecture #11
Chapter 10
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What Is this Course All About ?
• To Gain an Appreciation of the
Various Types of Signals and Systems
• To
T Analyze
A l
The
Th Various
V i
Types
T
off
Systems
• To Learn the Skills and Tools needed
to Perform These Analyses.
• To Understand How Computers
Process Signals and Systems
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Frequency Response of LTI
Systems
• Let’s
Let s review the Frequency Response for
continuous-time systems
• First,
First some definitions:
– The unit impulse function
– The
Th unit
i Step
S function
f
i
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A Special Function – Unit
Impulse Function
• The unit impulse function
function, δ(t),
δ(t) also known as
the Dirac delta function, is defined as:
δ(t) = 0 for t ≠ 0;
= undefined for t = 0
and has the following special property:
∞
∫ f (t)δ (t −τ )dt = f (τ )
−∞
∞
∴ ∫ δ (t)dt =1
−∞
δ(t)
0
-100
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-25
-1
0
1
25
50
100
274
Uses of Delta Function
• Modeling of electrical,
electrical mechanical,
mechanical physical
phenomenon:
–
–
–
–
point charge
charge,
impulsive force,
point mass
point light
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Unit Impulse Function
Continued
• A consequence of the delta function is that it
can be approximated by a narrow pulse as the
width of the pulse approaches zero while the
area under the curve = 1
δ(t)
lim δ (t) ≈1/ε for -ε / 2 < t < ε / 2; = 0 otherwise.
ε →0
10
1
0.5
-1
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.5
1
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Unit Impulse Function
Continued
∞
∫ f (t )δ (t − τ )dt
−∞
Let' s approximate δ (t − τ ) with a pulse of height
where ε = t-τ , we have
τ +ε 2
≈
∫
τ ε
− 2
1
ε
and width ε
1
f (t ) dt
ε
If we take the limit of this integral as ε → 0,
the approximate approaches the original integral
τ +ε 2
= lim
∫
ε → 0 τ −ε 2
1
1
f (t ) dt → lim f (τ ) ε → f (τ ),
ε
ε →0
ε
since as ε → 0, t → τ
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Another Special Function – Unit
Step Function
• The unit step function,
function u(t) is defined as:
u(t) = 1 for t ≥ 0;
1
= 0 for
f t < 0.
0
t
and is related to the delta function as follows:
t
−∞
u (t ) = ∫ δ (τ )dτ
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Integration of the Delta Function
• δ(t)
• u(t)
• tu(t)
u(t)
tu(t)
t 2 u(t)
.
.
.
•
2!
t n u((tt)
n!
1st order
2nd order
nth order
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Signal Representations using
the Unit Step Function
•
x(t) =
e-σ t
cos(ωt)u(t)
1.5
1
0.5
0
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
-0.5
-1
-1.5
•
x(t) = t u(t) – 2 (t-1)u(t-1) + (t-2) u(t-2)
4
3
2
1
0
-1 0
0.5
1
1.5
2
2.5
3
-2
-3
-4
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Convolution of the Unit-Impulse
Response
• As with Discrete
Discrete-time
time system, we find that the
Unit-Impulse Response of the a Continuous-time
system, h(t), is key to determining the output of
the system to any input:
y (t ) = h(t ) ⊗ x(t ) = ∫ h(τ ) x(t − τ )dτ
τ
• Let’s apply the complex exponential (sinusoidal)
signal, x(t)=Ae jφe jωt, for all t
y (t ) = h(t ) ⊗ x(t ) = ∫ h(τ ) x(t − τ )dτ
τ
= ∫ h(τ ) Ae jφ e jω ( t −τ ) dτ = {∫ h(τ )e − jωτ dτ } Ae jφ e jωt
τ
τ
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Frequency Response
• As in Discrete
Discrete-time
time systems, the Frequency
Response of the LTI system, H(jω), is:
y (t ) = {∫ h(τ )e − jωτ dτ } Ae jφ e jωt
τ
∞
H ( jω ) = ∫ h(t )e − jωt dt
−∞
y (t ) = H ( jω ) Ae jφ e jωt
• Again this is true only for sinusoidal signals!!!!
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Examples
•
Assume we have a system, H(j3π) = 2 - j2 with input x(t) = 10e j3ωt, then
y (t ) = H ( j 3π )10e j 3πt = (2 − j 2)10e j 3πt
= ( 2 2e
−j
= 20 2e
•
π
4
)(10e j 3πt )
π
j ( 3πt − j )
4
Assume h(t) = 2e-2tu(t), calculate the Frequency Response:
H ( jω ) =
∞
∫
−∞
=
−2 t
2e u (t )e
− jωt
∞
dt = ∫ 2e −2t − jωt dt
0
2
2
e − (2+ jω )t |∞0 =
[e − (2+ jω ) ∞ − e − (2− jω )0 ]
−(2 + jω )
−(2 + jω )
2
2[0e − jω∞ − 1
2
−2 ∞ − jω∞
[e e
=
− 1] =
=
−(2 + jω )
−(2 + jω ) 2 + jω
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Plotting the Frequency Response
• Let’s plot the magnitude
and
d phase
h
off the
th
Frequency Response vs.
Frequency
q
y
• What kind of filter is this?
Phase
H ( jω ) =
H ( jω ) =
2
2 + jω
2
22 + ω 2
ω
∠H ( jω ) = 0 − ∠{2 + jω} = − tan −1 ( )
2
Magnitude
1
1.5
1
0.5
0
-20
-10
0
10
20
-0.5
-1
0
-1.5
1.5
-20
20
-10
10
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10
20
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Frequency Response to a Cosine Input
• If the input
p to an LTI system
y
is
x(t)=A cos(ωt+Ԅ),
• and if the impulse response is real-valued, then the output will
be
y(t)=AM cos(ωt + Ԅ +φ)
• where the frequency response is
H(jω) = M e jφ
• To show this:
jωt }
x(t)
(t) =A
A cos(ωt+
( t+Ԅ) = ½A{e
½A{ jԄe jωt + e -jjԄe -jωt
• Using superposition
y(t)=
y(
) ½A{H(jω)
{ (j ) e jԄe jωt + H(-jω)
( j ) e -jԄe -jωt }
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Frequency Response to a Cosine
Input
• If the impulse response is real-valued then
y(t)= ½A{H(jω) e jԄe jωt + H*(jω) e -jԄe -jωt }
j e jԄe jωt
j t + (Me
j )* e -jjԄe -jωt
jψ
j t}
y(t)=
( ) ½A{
½A{Me jψ
(
y(t)= ½A{Me jψe jԄe jωt + Me -jψe -jԄe -jωt }
y(t)= ½A{Me j(ωt+Ԅ +ψ) + Me –j(ωt+Ԅ +ψ) }
y(t)=AM
y(t)
AM cos(ωt + +ψ)
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Proof of Conjugate Symmetry
*
∞
⎛
⎞
H * ( jω ) = ⎜⎜ ∫ h(t )e − jωt dt = ∫ h(t )*e + jωt dt
−∞
⎝ −∞
⎠
∞
∞
= ∫ h(t )e − j ( −ω )t dt = H (− jω )
−∞
h(t )* = h(t ) Only is h(t ) is real - valued
If H ( jω ) = Me jψ
Then H (− jω ) = H * ( jω ) = ( Me jψ )* = Me − jψ
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Examples
• Assume we have a system, given by the following H(jω)
and the input
p x(t)
( ) = 3 cos(40πt
(
– π)) is applied.
pp
Determine
the output signal.
40π
H ( jω ) =
40π + jω
40π
1
H ( j 40π ) =
=
2
(40π ) 2 + 40π 2
π
−1 40π
−1
) = − tan (1) = −
∠H ( j 40π ) = − tan (
40π
4
1
π
y (t ) = 3( ) cos(40πt − π − )
2
4
5π
= 2.1213 cos(40πt − )
4
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An Example
• An LTI system has an impulse response of
h(t) = δ (t) - 200πe-200πtu(t)
• The
h ffollowing
ll i signal
i l is
i applied:
li d
x(t)=10+20δ (t - 0.1)+40cos(200π t+0.3π) for all t
• The input has 3 parts: a constant, an impulse
and a cosine wave. We will take each part
p
separately and use the easiest method to find
the solution.
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An Example
• Let’s
Let s first find the frequency response of the system
from the impulse response:
∞
H ( jω ) = ∫ [δ (t ) − 200πe − 200πt u (t )]e − jωt dt
−∞
∞
= ∫ δ (t )e
− jω t
−∞
∞
dt − 200π ∫ e − 200πt u (t )e − jωt dt
−∞
∞
= 1 − 200π ∫ e −( 200π + jω )t dt = 1 +
0
= 1−
200π
e −( 200π + jω ) t |∞0
200π + jω
200π
jω
=
200π + jω 200π + jω
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An Example
•
Now let’s take the first (constant, ω = 0) part and the third (cosine) part and
evaluate the solution using the frequency response:
The
h first
fi part off the
h input
i
:10
H ( jω ) =
jω
200π + jω
j0
10 = 0
200π + j 0
The third part of the input : 40cos(200πt + 0.3π )
jω
H ( jω ) =
200π + jω
10 6 H ( j 0)10 =
40cos(200πt + 0.3π ) 6 40 H ( j 200π ) cos[200πt + 0.3π + ∠H ( j 200π )]
1∠
π
j 200π
j
2 = 1 ∠π
=
=
π
200π + j 200π 1 + j
2 4
2∠
4
1
40
40cos(200πt + 0.3
0 3π ) 6 40
cos[200πt + 0.3
0 3π + .25π ] =
cos[200πt + 0.55
0 55π ]
2
2
H ( j 200π ) =
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An Example
• Now for the second ppart of the input
p ((the impulse
p
function), we will apply the impulse response:
The second part of the input : 20δ (t − 0.1)
20δ (t − 0.1) 6 20h(t − 0.1) = 20[δ (t − 0.1) − 200πe − 200π ( t −0.1)u (t − 0.1)]
20δ (t − 0.1) 6 20δ (t − 0.1) − 4000πe − 200π (t −0.1)u (t − 0.1)
• The Complete solution by superposition is:
y (t ) = 0
+ 20δ (t − 0.1) − 4000π e −200π ( t −0.1)u (t − 0.1)
40
+
cos(200π t + 0.55
0 55π )
2
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Frequency Response to a Sum of
C i IInputs
Cosine
t
• We can extend this to the case when
x(t)=Σ Akcos(ωkt+Ԅk)
• By superposition
yk(t)=AM
( ) AMk cos(ω
( kt + Ԅk +φ
+ k)
• where the frequency response is
Hk(jωk) = Mk e jφk
• Then:
y(t) =Σ AMk cos(ωkt + Ԅk +φk)
• This model can also be used when analyzing periodic input
signals since a Fourier Series can be generated which has the
similar form: x(t)=Σ Akcos(kωot+Ԅk) where ωo is the
fundamental frequency.
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Homework
• Exercises:
– 10.1-10.3
• Problems:
– 10.1, 10.2,
– 10.4 Use Matlab to plot the frequency response and submit
your code
– Using unit step functions, construct a single pulse of
magnitude 10 starting at t=5 and ending at t=10.
– Repeat with
i h 2 pulses
l where
h the
h secondd is
i off magnitude
i d 5
starting at t=15 and ending at t=25.
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