Transcript Part 2

Frequency Response
Last time we
• Revisited formal definitions of linearity and time-invariance
• Found an eigenfunction for linear time-invariant systems
• Found the frequency response of a linear system to
eigenfunction input
• Found the frequency response for cascade, feedback,
difference equation, and differential equation systems
Today we will
• Extend the results to accommodate sinusoidal input, and
then any input via Fourier series representation
• Write the Fourier series in terms of complex exponentials
• Provide a method to calculate Fourier series coefficients
• Determine properties of these coefficients
EECS 20 Chapter 8 Part 2
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Response of an LTI System to Eigenfunction
Last time, we proved that for an input signal x given by
x(t) = eiωt for all t  Reals
the corresponding output y of an LTI system can be expressed as
y(t) = H(ω) eiωt for all t  Reals
where H(ω) is called the frequency response of the system.
The complex exponential eiωt is called an eigenfunction of the
system, because it creates an output with the same form, only
differing by a scaling factor.
The same is true for a discrete input,
x(t) = eiωn
for all n  Integers
y(n) = H(ω) eiωn for all n  Integers
EECS 20 Chapter 8 Part 2
leads to
2
Cosines as Complex Exponentials
Recall that cosines can be expressed as complex exponentials:
eit  eit
cos(t ) 
2
If we let x1(t) = eiωt and x2(t) = e-iωt, we express the cosine as
cos(ωt) = ½(x1(t) + x2(t))
If we apply the cosine as input to an LTI system S, we find
S(½(x1 + x2)) = ½ (S(x1) + S(x2))
and since x1 and x2 are eigenfunctions, we can write
y(t) = ½ (S(x1) + S(x2))(t)
= ½ (H(ω) eiωt + H(-ω) e-iωt )
So we can use frequency response to express the output for
sinusoidal input.
EECS 20 Chapter 8 Part 2
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Conjugate Symmetry
So for the input x(t) = cos(ωt), we obtain the output
y(t) = ½ (H(ω) eiωt + H(-ω) e-iωt ).
Realistic systems will produce purely real output (no imaginary
component) for a purely real input like cos(ωt).
This means that the imaginary parts of H(ω) eiωt and H(-ω) e-iωt
must cancel out; they must be opposite in sign.
This is the same as saying that one is the conjugate of the other:
H(ω) eiωt = (H(-ω) e-iωt)*
= H(-ω)* eiωt
For systems that produce real output for real input, it is true that
H(ω) = H(-ω)*
EECS 20 Chapter 8 Part 2
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Implications: Scaled and Shifted Sinusoids
Systems that produce purely real output for a purely real input
are called conjugate symmetric.
Let’s look again at the output for our case x(t) = cos(ωt),
y(t) = ½ (H(ω) eiωt + H(-ω) e-iωt )
Using the fact that z + z* = 2 Re{z},
y(t) = ½ (2 Re{H(ω) eiωt })
= Re{H(ω) eiωt }

If we express H(ω) in polar form, H(ω) = |H(ω)| ei H(ω),

i
y(t) = Re{|H(ω)| e H(ω) eiωt }

= Re{|H(ω)| ei( H(ω)+ωt) }
EECS 20 Chapter 8 Part 2
= |H(ω)| cos(ωt+H(ω))
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Computing Sinusoidal Response
So, given the system response to an eigenfunction, H(ω), we can
compute the magnitude response |H(ω)| and the phase
response H(ω).
These form the scaling factor and phase shift in the output,
respectively.
The frequency of the output sinusoid will be the same as
the frequency of the input sinusoid in any LTI system.
The LTI system scales and shifts sinusoids.
These results hold true for both continuous and discrete signals
and systems.
EECS 20 Chapter 8 Part 2
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Example
Consider our RC circuit from last time,
where we found
1
H() 
1  iRC
x(t)
+
_
R
+
y(t)
_
C
To compute the voltage over the capacitor, y(t), for a sinusoidal
input voltage x(t), I simply need to find the magnitude and
phase of H(ω) and plug in:
1
1
| H() | 

| 1  iRC |
1  (RC)2
H()  1  (1  iRC)   tan1 RC
EECS 20 Chapter 8 Part 2
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Response to Fourier Series Input
In Chapter 7, we mentioned that any periodic signal can be
represented by a Fourier series:

x( t )  A 0   A k cos(k0 t  k )
k 1
Since we are dealing with LTI systems, where we can pull out
constants and distribute over sums, we can get the system
output for any input by scaling and summing the output for the
individual sinusoids in the Fourier series.
EECS 20 Chapter 8 Part 2
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Alternate Fourier Series Representation
Remembering that
eit  eit
cos(t ) 
2
we may write


 A
x( t )  A 0   k ei(k0 t  k )  e i(k0 t  k )
k 1 2
 A
A
x( t )  A 0   k eik eik0 t  k e ik e ik0 t
2
k 1 2
A
 0
Xk   12 A k eik
1
i
 2 A k e k
we obtain
EECS 20 Chapter 8 Part 2
if k  0
and also
and letting

ik0 t
ik0 t
x
(
t
)

X

X
e

X
e
0  k
k
if k  0
if k  0
k 1
or simply

x( t )   Xk eik0 t
k  
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Alternate Fourier Series Representation: Discrete
For a discrete periodic signal, with the new notation
A0



 12 A k eik

Xk  
 A cos( )
k
 k

ip k
1 A
e
 2 p k
if k  0
 
if k  p 2
 
x(n) 
if k  p 2
p 1

k 0
Xk eik0n
 
if k  p 2
The proof is given in the text on page 303.
EECS 20 Chapter 8 Part 2
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Response to Fourier Series Input
Now let’s apply a continuous input x(t) to an LTI system with
frequency response H(ω) and find the output y(t):

x( t )   Xk eik0 t
k  
Due to linearity, we can distribute over the sum and pull out the
constants Xk.
The result is a scaled sum of the output generated by each
̥
ikω
individual complex exponential e t.
Since each eiωt has corresponding output H(ω)eiωt ,

y( t )   XkH(k0 )eik0 t
k  
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Determining Fourier Series Coefficients
We now give formulae for the Fourier series coefficients for a
periodic signal of period p:
p
For continuous signals
x(t), t  Reals:
1
Xm   x( t )e im 0 t dt
p0
For discrete signals x(n)
x(n), n  Integers:
1 p 1
Xm   x(n)e im0n
p n 0
The textbook provides a validation of these formulae on page
306, but their derivation will be intuitive once we have covered
Fourier transforms.
EECS 20 Chapter 8 Part 2
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