Frequency Response
Lecture #12
Chapter 10
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Ideal Filters
• We want to study H(jω) functions which
provide frequency selectivity such as:
– Low Pass
– High Pass
– Band Pass
• However, we will look a ideal filtering, that is,
filter which have ideal performance but are
very difficult to construct.
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A simple Filter – Ideal Delay
•
•
•
Ideal Delay Filter => y(t) = x(t – td): the output is same as the input
except shifted in time by an amount td seconds.
The impulse response is just h(t) = δ(t – td)
The frequency response is then




H ( j )   h(t )e  jt dt    (t  t d )e  jt dt
 e  jtd
Alternatively, if x(t )  e j (t  ) , then
y (t )  x(t  t d )  e j[ (t td )  ]  e  jtd e j (t  )
But y (t )  H ( j )e j (t  )
And therefore,
H ( j )  e  jtd
•
•
The Frequency Response of an Ideal Delay filter has a constant
magnitude with a phase that is linear with frequency
Therefore, it does not affect the magnitude of the input. It only effects the
phase by an amount of -ωtd
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Example
A signal of the form x(t )  10e j 4 e j 200t is input to an ideal
delay filter with delay of 0.001 sec.
The frequency response is : H ( j )  e  j 0.001
H ( j 200 )  e  j 200 ( 0.001)
Then the output signal becomes :
y (t )  H ( j ) x(t )  H ( j 200 )10e j 4 e j 200t
 e  j 200 ( 0.001)10e j 4 e j 200t  e  j 0.2 10e j 4 e j 200t
 10e

j ( 200t   0.2 )
4
 10e j ( 200t  0.05 )
Or rewritten as : y (t )  10e

j [ 200 ( t  0.001)  ]
4
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Ideal Low Pass Filter
• This filter only passes frequencies below a value ωco and
attenuates all frequencies above ωco.
• We call ωco the cutoff frequency.
• Therefore, the frequency response of a low pass filter is:
1
  co
0
  co
H lp ( j ) 
ωco
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ωco
301
Ideal High Pass Filter
• This filter only passes frequencies above a value ωco and
attenuates all frequencies below ωco.
• We call ωco the cutoff frequency.
• Therefore, the frequency response of a low pass filter is:
1
  co
0
  co
H hp ( j ) 
ωco
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ωco
302
Ideal Band Pass Filter
• This filter only passes frequencies above a value ωco1 and
below a value ωco2 and attenuates all other frequencies
outside this range.
• We call ωco1 the lower (or low) cutoff frequency and
ωco2 the upper (or high) cutoff frequency.
• Therefore, the frequency response of a low pass filter is:
0
H bp ( j ) 
1
0
  co1
co1    co 2
  co 2
BME 310 Biomedical Computing J.Schesser
-ωco2 ωco1
ωco1 ωco2
303
Application of Ideal Filters
•
•
We will apply a band pass filter to a periodic
square wave filter out its fundamental
frequency.
Let our input signal have a period of
To = 500µs or fo=2kHz => ωo=2π(2000) rad/sec
and its form over one period is:
x(t)
2 0 ≤ t < To / 2
x(t) =
0 To / 2 ≤ t < To
2
-2To
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-To –To/2 0
To/2 To
2To
304
t
Application of Ideal Filters
• Since x(t) is a period, let’s calculate the Fourier
series for to decompose the input into its
frequency components.
x(t ) 

a e
k  
1
ak 
To
k
jko t
To 2
 x(t )e
 jk o t
To 2
ak 
1
dt 
To
To 2
 2e
 jko t
dt
2 To
2

T 2
1 o
1
x(t)dt 
a0 

To To 2
To
0
 jk
2
1
 jk o t To 2
e
|0 
[e To

To ( jko )
 jk
1
1
2
2  j /2
[1 e  jk ] 
[1  (1)k ] 

e
for odd values of k
jk
jk
jk k
0
for even values of k
e
 jk
2
0
To
To 2

2 dt  1
0
]
1
1
[1  e  jk ]
jk
Recall that e  jk  cos k  j sin k  1 for even values of k
 1 for odd values of k
or e
 jk
 (1)
k
0
-4
-3
-2
-1
0
1
2
3
4
k
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Application of Ideal Filters
• Now let’s apply an ideal band pass filter with low
frequency cutoff of 1,250 Hz and high frequency cutoff
of 2,750 Hz which has a bandwidth of 1500 Hz and is
centered around 2000 Hz which is the fundamental
frequency of this square wave.
1
0
-12
-10
-8
-6
-4
-2
0
2
4
6
8
10
12
Hz
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Application of Ideal Filters
• If the filter is LTI, then the output signal is also
periodic with same fundamental frequency.
Therefore, y(t) can be written as a Fourier Series.
y(t) 

be
k
jk ot
k
By superposition of these complex exponential signals
at k o
y(t)  b1e j 2  (2000)t  b1e  j 2  (2000)t
2  j /2 j 2  (2000)t
2
e
e

e  j /2 e  j 2  (2000)t )
 (1)
 (1)
2
 ( e j(2  (2000)t /2)  e  j(2  (2000)t /2) )

bk  H ( jk o )ak
But since H ( j ) is only defined for 1250    2750,

then only the terms for which k  1 will be left

upon this multiplication
2  j /2
e
for k  1
bk  H ( jk o )ak 
k
BME 310 Biomedical Computing J.Schesser


2

2

4

( e j(2  (2000)t /2)  e j e  j(2  (2000)t /2) )
( e j(2  (2000)t /2)  e  j(2  (2000)t /2) )
cos(2 (2000)t   / 2)
307
Time Domain or Frequency
Domain
• We have seen that a LTI can be represented by its
impulse response in the time domain and by its
frequency response in the frequency domain.
• In general when working with sinusoids (or complex
exponentials) either single or summed signals, it is
easier to work in the Frequency Domain.
• If the signal consists of impulses, step functions, or
other non-sinusoidal signals (e.g., signals which are
progressive integrations of the impulse function),
convolution of the impulse response (Time Domain)
is usually easiest.
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An Example
• An LTI system has an impulse response of
h(t) = δ (t) - 200πe-200πtu(t)
• The following signal is applied:
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
separately and use the easiest method to find
the solution.
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An Example
• Let’s first find the frequency response of the system
from the impulse response:

H ( j )   [ (t )  200e  200t u (t )]e  jt dt


   (t )e
 j t


dt  200  e  200t u (t )e  jt dt


 1  200  e ( 200  j )t dt  1 
0
 1
200
e ( 200  j )t |0
200  j
200
j

200  j 200  j
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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 first part of the input :10
j
H ( j ) 
200  j
j0
10  H ( j 0)10 
10  0
200  j 0
The third part of the input : 40cos(200t  0.3 )
j
H ( j ) 
200  j
40cos(200t  0.3 )  40 H ( j 200 ) cos[200t  0.3  H ( j 200 )]
1

j 200
j
2  1 



200  j 200 1  j
2 4
2
4
1
40
40cos(200t  0.3 )  40
cos[200t  0.3  .25 ] 
cos[200t  0.55 ]
2
2
H ( j 200 ) 
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An Example
• Now for the second part of the input (the impulse
function), we will apply the impulse response:
The second part of the input : 20 (t  0.1)
20 (t  0.1)  20h(t  0.1)  20[ (t  0.1)  200e  200 (t 0.1)u (t  0.1)]
20 (t  0.1)  20 (t  0.1)  4000e  200 (t 0.1)u (t  0.1)
• The Complete solution by superposition is:
y (t )  0
 20 (t  0.1)  4000 e 200 ( t 0.1)u (t  0.1)
40

cos(200 t  0.55 )
2
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Homework
• Exercises:
– 10.4-10.7
• Problems:
– 10.5, 10.6,
– 10.7 Use Matlab to plot x(t); show your code
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