ECE
8443
– PatternContinuous
Recognition
EE
3512
– Signals:
and Discrete
LECTURE 16: FOURIER ANALYSIS OF CT SYSTEMS
• Objectives:
Response to a Sinusoidal Input
Frequency Analysis of an RC Circuit
Response to Periodic Inputs
Response to Nonperiodic Inputs
Analysis of Ideal Filters
• Resources:
Wiki: The RC Circuit
CN: Response of an RC Circuit
CNX: Ideal Filters
URL:
Differential Equations
• For CT systems, such as
circuits, our principal tool is
the differential equation.
• For the circuit shown, we can
easily compute the
input/output differential
equation using Kirchoff’s Law.
Ri(t )  y (t )  x(t )  0
dvC (t )
dy(t )
C
dt
dt
dy(t )
RC
 y (t )  x(t )  0
dt
dy(t )
1
1

y (t ) 
x(t )
dt
RC
RC
i (t )  C
EE 3512: Lecture 16, Slide 1
• What is the nature of the impulse
response for this circuit?
Numerical Solutions to Differential Equations
• Consider our 1st-order diff. eq.:
• We can replace n by n-1 to obtain:
dy (t )
y[n]  (1  aT ) y[n  1]  bTx[n  1]
 ay (t )  bx (t )
dt
• This is called the Euler approximation
• We can solve this numerically by
to the differential equation.
setting t = nT:
dy(t )
• With x[n]  0 n and initial condition,
 ay(nT )  bx(nT )
y[0] , the solution is:
dt t nT
• The derivative can be approximated:
y[n]  (1  aT) n y[0], n  0, 1, 2, ...
dy(t )
y(nT  T )  y(nT )
• The CT solution is:

dt t nT
T
 at
y
(
t
)

e
y(0), t  0
• Substituting into our diff. eq.:
y (nT  T )  y (nT )
• Later, we will see that using the
 ay (nT )  bx (nT )
T
Laplace transform, we can obtain:
• Let x(nT )  x[n] and y(nT )  y[n] :
y[n]  e anT y[0], n  0, 1, 2, ...
y[n  1]  y[n]
 ay[n]  bx[n]
T
y[n  1]  y[n]  aTy[n]  bTx[n]
y[n  1]  (1  aT ) y[n]  bTx[n]
EE 3512: Lecture 16, Slide 2
• But we can approximate this:
a 2T 2 a 3T 3
 aT
e
 1  aT 

 ...
2
6
• Which tells us our 1st-order
approximation is accurate!
Higher-Order Derivatives
• We can use the same approach for the second-order derivative:
d 2 y (t )

dt t nT

dy(t )
dt

t  nnT  T
dy(t )
dt
t  nnT
T
y (nT  2T )  2 y (nT  T )  y (nT )
T
• Higher-order derivatives can be similarly approximated.
• Arbitrary differential equations can be converted to difference equations using
this technique.
• There are many ways to approximate derivatives and to numerically solve
differential equations. MATLAB supports both symbolic and numerical
solutions.
• Derivatives are quite tricky to compute for discrete-time signals. However, in
addition to the differences method shown above, there are powerful methods
for approximating them using statistical regression.
• Later in the course we will consider the implications of differentiation in the
frequency domain.
EE 3512: Lecture 16, Slide 3
Series RC Circuit Example
dy (t )
1
1

y (t ) 
x(t )
dt
RC
RC
Difference Equation:
R=1;C=1;T=0.2;
a=-(1-T/R/C);b=[0 T/R/C];
y0=0; x0=1;
n=1:40;
x=ones(1,length(n));
y1=recur(a, b, n, x, x0, y0);
Analytic Solution:
t=0:0.04:8;
y2=1-exp(-t);
y1=[y0 y1];
n=0:40;
plot(n*T, y1, ’o’, t, y2, ’-’);
EE 3512: Lecture 16, Slide 4
Example: RC Circuit
dy (t )
1
1

y (t ) 
x(t )
dt
RC
RC
• Using our FT properties:
jY   
1
1
Y   
X ( )
RC
RC
1 / RC
Y ( ) 
X  
j  1 / RC
Y ( )
1 / RC
H ( ) 

X   j  1 / RC
1 / RC
H ( ) 
 2  (1 / RC) 2
H ( )   t an1 RC 
• Compute the frequency response:
RC = 0.001;
W=0:50:5000;
H=(1/RC)./(j*w+1/RC);
magH=abs(H);
angH=180*angle(H)/pi;
EE 3512: Lecture 16, Slide 5
Response of an LTI System to a Sinusoid
• Consider an LTI CT system with impulse response h(t):

y(t )  h(t ) * x(t )   h( ) x(t   )d

• We will assume that the Fourier transform of h(t) exists:

H ( )   h(t )e  jt dt

• The output can be computed using our Fourier transform properties:
Y ()  H () X () and Y ()  H () X () Y ()  H ()  X ()
• Suppose the input is a sinusoid:
x(t )  A cos(0 t   )
• Using properties of the Fourier transform, we can compute the output:


X ( )  A e  j     0   e j     0 
Y ( )  H ( ) X ( )


 AH ( ) e  j     0   e j     0 


 A H ( 0 )e  j     0   H ( 0 )e  j     0 

 A H ( ) e

 A H ( 0 ) e  j H (0 ) e  j     0   e j H (0 ) e j     0 
0
    0   e j H (
 j H (0 )  
0 ) 
y (t )  F -1 Y ( )   A H ( 0 ) cos 0 t    H ( 0 ) 
EE 3512: Lecture 16, Slide 6

    0 

Example: RC Circuit (Cont.)
• We can compute the output for RC=0.001 and ω0=1000 rad/sec:
y(t )  A(0.707) cos1000t    45
• We can compute the output for RC=0.001 and ω0=3000 rad/sec:
y(t )  A(0.316) cos3000t    71.6
• Hence the circuit acts as a lowpass filter. Note the phase is not linear.
• If the input was the sum of two sinewaves:
x(t )  cos100t   cos3000t 
describe the output.
EE 3512: Lecture 16, Slide 7
Response To Periodic Inputs
• We can extend our example to all periodic signals using the Fourier series:

x(t )  a0   Ak cosk0 t   k  (a variantof the trigonometric Fourier series)
k 1
• The output of an LTI system is:

y(t )  a0 H 0   Ak H k0  cosk0 t   k  H k0 
k 1
• We can write the Fourier series for the output as:


y (t )  a   Aky cos k 0 t   ky
y
0

k 1
where,
a 0y  a 0x H (0)
Aky  Akx H (k 0 )  ky   kx  H (k 0 )
also,
1 x
Ak H (k 0 ) and c ky   kx  H (k 0 )
2
• It is important to observe that since the spectrum of a periodic signal is a line
spectrum, the output spectrum is simply a weighted version of the input,
where the weights are found by sampling of the frequency response of the
LTI system at multiples of the fundamental frequency, 0.
c ky 
EE 3512: Lecture 16, Slide 8
Example: Rectangular Pulse Train and an RC Circuit
• Recall the Fourier series for
a periodic rectangular pulse:

x(t )  a0   a k coskt 
k 1
where,
sin k / 2
(k / 2)
• Also recall the system response was: H ( ) 
a 0  0.5 a k 
• The output can be easily written as:


y (t )  a   Aky cos k 0 t   ky
y
0

k 1
where,
a0y  a0x H (0)  0.5
Aky  Akx H (k 0 ) 
 2

  k


sin k / 2
1 / RC
(k / 2)
(k ) 2  (1 / RC) 2
1 / RC
(k )  (1 / RC)
0
2
EE 3512: Lecture 16, Slide 9
2
k odd
k even
1 / RC
j  1 / RC
Example: Rectangular Pulse Train (Cont.)
• We can write a similar expression for the output:
y(t )  a 
y
0

2
 k
k 1
k odd
1 / RC
(k ) 2  (1 / RC) 2
coskt  tan1 kRC 
1/RC = 1
• We can observe the implications of
lowpass filtering this signal.
• What aspects of the input signal give
rise to high frequency components?
1/RC = 10
• What are the implications of
increasing 1/RC in the circuit?
• Why are the pulses increasingly
rounded for lower values of 1/RC?
• What causes the oscillations in the
signal as 1/RC is increased?
EE 3512: Lecture 16, Slide 10
1/RC = 100
Response to Nonperiodic Inputs
• We can recover the output in the time domain using the inverse transform:

1
j
j
jt
y(t ) 
H
(
e
)
X
(
e
)
e
d

2 
• These integrals are often hard to compute, so we try to circumvent them
using transform tables and combinations of transform properties.
• Consider the response of our RC circuit to a single pulse:
sin( / 2)
( / 2)
1 / RC
H ( e j ) 
j  1 / RC
X ( e j ) 
Y (e j )  X (e j ) H (e j )
sin( / 2) 1 / RC

( / 2) j  1 / RC
• MATLAB code for the frequency response:
RC=1;
w=-40:.3:40;
X=2*sin(w/2)./w;
H=(1/RC)./(j*w+1/RC);
Y=X.*H;
magY=abs(Y);
EE 3512: Lecture 16, Slide 11
Response to Nonperiodic Inputs (Cont.)
• We can recover the output using the inverse Fourier transform:
syms X H Y y w
X = 2*sin(w/2)./w;
H=(1/RC)./(j*w+1/RC);
Y=X.*H;
Y=ifourier(Y);
ezplot(y,[-1 5]);
axis([-1 5 0 1.5])
1/RC = 1
1/RC = 1
1/RC = 10
EE 3512: Lecture 16, Slide 12
1/RC = 10
Ideal Filters
• The process of rejecting particular frequencies or a range of frequencies is
called filtering. A system that has this characteristic is called a filter.
• An ideal filter is a filter whose frequency response goes exactly to zero for
some frequencies and whose magnitude response is exactly one for other
ranges of frequencies.
• To avoid phase distortion in the filtering process, an ideal filter should have a
linear phase characteristic. Why?
H (e j )  t d for all ω in the filter passband
• We will see this “ideal” response has some important implications for the
impulse response of the filter.
• Lowpass
• Highpass
• Bandpass
• Bandstop
EE 3512: Lecture 16, Slide 13
Ideal Linear Phase Lowpass Filter
• Consider the ideal lowpass filter
with frequency response:
• Phase
Response
e  jtd ,  B    B
H (e )  
  B,   B
 0
j
• Using the Fourier transform pair
for a rectangular pulse, and applying
the time-shift property:
h(t ) 
B

sin c (t  t d 



B
• Is this filter causal?
• The frequency response of an ideal
bandpass filter can be similarly defined:
e  jtd , B1    B2
H (e )  
elsewhere
 0,
j
• Will this filter be physically realizable?
Why?
EE 3512: Lecture 16, Slide 14
• Impulse
Response
Summary
• Showed that the response of a linear LTI system to a sinusoid is a sinusoid at
the same frequency with a different amplitude and phase.
• Demonstrated how to compute the change in amplitude and phase using the
system’s Fourier transform.
• Demonstrated this for a simple RC circuit.
• Generalized this to periodic and nonperiodic signals.
• Worked examples involving a periodic pulse train and a single pulse.
• Introduced the concept of an ideal filter and discussed several types of ideal
filters.
• Noted that the ideal filter is a noncausal system and is not physically
realizable. However, there are many ways to approximate ideal filters, and that
is a topic known as filter design.
EE 3512: Lecture 16, Slide 15