The sampling of continuous-time signals is an important topic It is required by many important technologies such as: Digital Communication Systems ( Wireless Mobile Phones, Digital TV (Coming), Digital Radio etc ) CD and DVD Digital Photos

Switch close and open Periodically with period T s Discrete Level If you have 8 levels you will need 3 bits If you have 16 levels you will need 4 bits Analog or continues level

Commercial type ADC or A/D

Some application of ADC or A/D

Recall Fourier Transform of periodical signal

Fourier Transform of periodical signal Generating function

Since

Let Maximum Frequency in F(  ) Ideal Sampling in time

Question : How to recover the original signal f(t) ↔F(  ) from the sampled F s (  ) Ideal Sampling in time

Question : How to recover the original continuous signal f(t) ↔F(  ) from the sampled f s (t) ↔ F s (  ) Ideal Sampling in time If I can recover this in frequency With a constant correction A/Ts → A I can Fourier inverse back to recover the original continuous signal f(t) If I use ideal low pass filter I will be able to extract this

If the bandwidth of the ideal low pass filter is greater than We will get distorted shape When we inverse back we will not get the original signal f(t) Therefore the ideal low pass filter bandwidth should be

Ideal Sampling in time Now what will happened if you lowered the sampling frequency The frequencies from adjacent part of the spectrum will interfere with each other We get distortionAliasing مستعار

Therefore to avoid aliasing and recover the original signal the sampling should be such that Therefore Nyquist proposed the following The sampling rate (  s) must be at least twice the highest frequency (  B) component present in the sample in order to reconstruct the original signal. The Sampling Theorem

Discrete Time Discrete Level

Switch close and open Periodically with period T s Discrete Level Analog or continues level We know have the following definition Therefore we will have a sequence of numbers Next we develop the mathematics for discrete signals

Note that the discrete-time impulse function is well behaved mathematically and presents none of the problems of the continuous time impulse function The shifted unit impulse function is defined by

The discrete-time unit impulse function can be expressed as the difference of two step functions

10 Discrete-Time Linear Time-Invariant Systems Recall from the continuous case

Impulse Input Impulse response Shifted Impulse InputShifted Impulse Response Linear –Time Invariant Multiply by constant Multiply by the response by the same constant An equation relating the output of a discrete LTI system to its input will now be developed Recall from the continuous case

Now relating the output of a discrete LTI system to its input will now be developed Multiply each side by x[k]

(Discrete Convolution) Let the input to a discrete-time system and the unit impulse response

Consider the system

Recall from the continuous case Recall that a memory less (static) system is one whose current value of output depends on only the current value of input. A system with memory is called a dynamic system

A discrete-time LTI system is causal if the current value of the output depends on only the current value and past values of the input Recall from the continuous case

Z-Transform of the Sequence samples x(nT) ≡ x[n] The coefficient x[n] denotes the sample values and z  n denote the Sample occurs n sample periods after t = 0

Define the unite impulse sequence by, Note : the unit impulse here (the discrete) is different from the impulse  (t) LaplaceZ

Define the unite step by the sample values

Z- Transform Properties (1) LinearityZ- Transform is Linear operator Then Proof

Define the unite step by the sample values

Proof : see the book page 549

We want the z-Transform of cos(bn ) Entry on the Table 11.2 Similarly

Let the input to a discrete-time system and the unit impulse response Using the same procedures we used in Fourier Transform and Laplace Transform we get The transfer function

Using partial fraction One problem occurs in the use of the partial-fraction expansion procedure of Appendix F The numerator is constant However Z-transform for the exponential Has z variable in the numerator To solve this problem we expand (partial fraction ) Y(z)/z

Using the Z -Transform Table Find x(n) ? Since the degree of the Numerator equal the degree of the denominator Polynomial division This form is not available on the table

Now the degree of the Numerator less than the degree of the denominator Using partial fraction, we have Using Table 11-1

Inverse Z -Transform Since Therefore, if we can put X(z) into the form shown above, Then we can determine x(nT) by inspection x(nT) will be the coefficients of the polynomial of X(z)

Using polynomial division, we get Therefore The disadvantage of this method is that, we do not get x(nT) in closed form