2. 1 Conditions for Distortionless Transmission Transmission is said to be distortion less if the input and output have identical wave shapes within a multiplicative constant. Transmission is said to be distortion less if the input and output have identical wave shapes within a multiplicative constant. A delayed output that retains the input waveform is also considered distortion less. A delayed output that retains the input waveform is also considered distortion less. Thus in distortion-less transmission, the input x(t) and output y(t) satisfy the condition: y(t) = Kx(t -  ) (1) where  is the delay time and k is a constant. where  is the delay time and k is a constant. Computing the Fourier Transform of (1) we obtain Computing the Fourier Transform of (1) we obtain Y(w) = KX(w)e -jw  (2) Y(w) = KX(w)e -jw  (2) The magnitude and phase response of (2) is given by The magnitude and phase response of (2) is given by

3. 2 |H(w)| = K and  (w) = -w  = -2  f  |H(w)| = K and  (w) = -w  = -2  f  These are plotted in the following figure. K |H(w)|  (w) ww  -w  Amplitude response Phase response A physical transmission system may have amplitude and phase responses such as those shown below:  (w) |H(w)| w w

4. 3 Ideal Filters Filter: A very general term denoting any system whose output is a specified function of its input. Frequency Selective Filters: Low-Pass, High-Pass, Band-Pass, Band-Stop. Ideal Low-Pass Filter: An ideal low-pass filter passes all Signal components having frequency less than w w radian per second with no distortion and completely attenuates signal components having frequencies greater than w c Hz. -w c w c w |H(w)|  (w) w

5. 4 Ideal High-Pass Filter: An ideal High-Pass filter passes all signal components greater than w w radian per second with no distortion and completely attenuates signal components having frequencies less than w w radian per second. |H(w)| -w c wcwc w w  (w)

6. 5 Ideal Band pass Filter: An ideal Band stop filter passes all signal components having frequencies in a band of B centered at the frequency w 0 with no distortion and completely attenuates signal components having frequencies outside this band. -w 0 w0w0 B

7. 6 Ideal Band stop Filter: An ideal Band stop filter is defined in the following figure: |H(w)|  (w) w w

8. 7 Characteristics of Practical Frequency Selective Filters 1+  1 1-  1 Passband ripple  1 = passband ripple  2 = Stopband ripple w c = w p = passband edge frequency. w s = stopband edge frequency. wpwp wsws 22

9. 8 Analogue Filters: The Low-Pass Butterworth Approximation: The Low-Pass Butterworth Approximation: A Low-pass Butterworth filter has the amplitude response where n  1 is the filter order and the subscript b denotes the Butterworth filter. w c is the cutt-off frequency of the filter. (1) It is obvious from equation (1) that the Butterworth filter is an all Pole filter (i.e. N poles but no zeros).

10. 9 02468w 0 1 N = 1 N=2 N=3 N=4 |H b (w)| The magnitude response of a Butterworth filter of order 1, 2, 3 and 4. Cutt-off Frequency is 1 radian per second.

11. 10 or The poles of the filter are the roots of the denominator, i.e. or k = 0,1,2,…., N-1 The poles of a Butterworth filter can be computed as follows: From (1) (2)

12. 11 Example1: Derive the transfer function of a first-order Butterworth filter. The cut-off frequency is 1 radian per second. Example1: Derive the transfer function of a first-order Butterworth filter. The cut-off frequency is 1 radian per second. Solution: The poles of a first-order Butterworth filter can be computed by putting k=0 and N = 1 in equation (2). i.e. s 0 = w c e j  /2 e j  /2 = e j  (w c = 1) = cos  + jsin  = -1 + 0 = -1 = cos  + jsin  = -1 + 0 = -1 This means that the transfer function of the filter is

13. 12 Example2: Repeat example 1 for a second order Butterworth filter. Example2: Repeat example 1 for a second order Butterworth filter. Solution: The poles of a second-order Butterworth filter can be computed by putting k=0, 1 and N = 2 in equation (2). i.e. s 0 = w c e j  /2 e j  /4 = e j 3  /4 (w c = 1) = cos(3  /4) + jsin(3  /4) = -1/  2 + j1/  2 = cos(3  /4) + jsin(3  /4) = -1/  2 + j1/  2 and s 1 = e j  /2 e j3  /4 = e j5  /4 = -1/  2 - j1/  2 This means that the transfer function of the filter is Tutorial: Repeat example 2 for a 3 rd and 4 th order Butterworth filter.

14. 13 Chebyshev Filter: There are two types of Chebyshev filters: Type1 Chebyshev Filters: These are all pole filters that Exhibit equi-ripple behaviour in the passband and a Monotonic characteristic in the stop band, as shown in the following figure. 0 1 wpwp w 1/(1+  2 )

15. 14 Type2 Chebyshev Filter: These filters contain both poles and zeros and exhibit a monotonic behaviour in the passband and an equiripple behaviour in the stopband. The magnitude response of a typical low-pass type 2 chebyshev filter is shown in the following figure. 0246810 0 1

16. 15 The magnitude of the frequency response characteristics of a type1 Chebyshev filter is given by where  is a parameter of the filter that is related to the ripple in the pass-band and T N (x) is the Nth order Chebyshev polynomial defined as The Chebyshev polynomials can be generated by the recursive equation T N+1 (x) = 2xT N (x) – T N-1 (x), N = 1,2,… (3) where T 0 (x) = 1 and T 1 (x) = x. From (3) T 2 (x) = 2x 2 – 1, T 3 (x) = 4x 3 – 3x, and so on.

17. 16 The filter parameter  is related to the ripple in the passband, as shown in the figure of the previous slide. A relationship between passband ripple  1 and the parameter  is given by A relationship between passband ripple  1 and the parameter  is given by  1 = 10log(1 +  2 )  1 = 10log(1 +  2 ) or  =  (10  1/10 – 1) or  =  (10  1/10 – 1) Example3: Derive transfer function of a first-order Cheby- Shev filter of type 1 with a unity gain and a passband ripple of 2dB. Solution:  =  (10 2/10 –1) = 0.7648, T 1 2 = (w/w c ) 2

18. 17 Therefore, Example 4: Find the transfer function for a second order normalized (w c = 1) Chebyshev low-pass filter with unity maximum gain and 1.5 dB of ripple in the passband. Solution:  1 = 1.5 dB, w c = 1,  2 = 10 1.5/10 – 1 = 0.4125

19. 18 Tutorial Q2: Derive the transfer function of a second order Low-pass chebyshev filter with unity dc gain and a passband Ripple of 2dB.