Discrete Time Systems

Introduction

y(nTs)
 
In our discussion of Sampling we introduced an analog system which was equivalent to a discrete-time system.

 

 

 


If we simplify our notation,

and

xn
 

yn
 

hn
 
We get the discrete-time system.

 

 


Where the input, output and “pulse response” are each sequences of sampled values of the corresponding analog function.

Macroscopic Time Domain Analysis

In our Linear Systems discussion, we found that the output, y(t), of a linear system is related to its input, x(t), by the Convolution Integral,

   Note: the limits normally are from 0 to t since the input starts at t=0 and the impulse response is 0 for values less than 0

But, in this case,

    For x(t) = 0 when t < 0 and assuming ideal sampling

Therefore

Now interchanging the order of integration and summation and using the sifting property of d-functions

And using the sifting property of the delta function

Now we resample at t = n*Ts:

Again using our simplified notation, we obtain the “Convolution Sum”

Where hn is the sequence of samples of the original impulse response at the t = n*Ts

If we let the input sequence, xn, be the unit pulse at n = 0

                  

             so hn is the “Unit Pulse Response” of the discrete-time system

In general h(n,k) is the response at sample time n due to a pulse that occurred at sample time k

An example discrete-time system

 

yn
 
 

 

 

 

 


Let xn = dn Therefore yn = { 1, a, a2, a3, …..) = an and hn = an for n ≥ 0

Note:   yn = 0 for n < 0 and for stability | a| ≤ 1
since this discrete-time system is equivalent (within a gain factor) to the simple RC low-pass filter (“leaky bucket”) which has been used often as an analog system example in these notes.  The time Constant, RC, is:

           

We can now solve for the output of this system given any input through the use of the convolution sum.

Let xn be the discrete-time version of the unit step function

           

And let the system be initially at rest (yn = 0).

The output is given by the convolution sum as

 The second form is the commutative version and is easier to use for this calculation.

We have two cases:

For n < 0 all of the input values are zero and the sum is zero

For n ≥ 0 the sum becomes

           

This is a simple geometric series.  To get a closed form without resorting to formulas pull out the first term and add/subtract an extra term

           

Now let p = k - 1

           

           

             or

             for n ≥ 0

We therefore have our macroscopic approach to solving for the output of discrete-time systems.

Microscopic Time Domain analysis

For a microscopic analysis note how we determined hn for the example system.  We actually calculated each output at time n from a knowledge of the input at time n and the “state” of the memory (or delay elements) in the system.  For the example system:

             or

           

This is a finite difference equation which describes the example system at a moment in time.  It is analogous to the differential equation:

           

which describes the RC low pass example in the discussion of Linear Systems.

These difference equations can be solved directly using the same techniques used to solve differential equations.  As an example we will solve the same problem used in the section on discrete convolution.

            ; 0 < a < 1; xn is the unit step and the system is initially at rest with yn = 0.

Step 1: Find the homogeneous (AKA natural or transient) solution

           

            Assume an exponential solution yn = Cn

            Cn  - aC(n-1) = 0

            C n  = aC(n-1) or

                  C = a

The homogeneous solution is therefore

            yn = KH * an

Step 2: Find the particular (AKA forced or steady-state) solution – we shall again use the method of undetermined coefficients

            The particular solution for this problem must be of the form

            yn = K0 + K1*n            since the input is a constant. Therefore

            y(n-1) = (K0 – K1) +K1*n

            Now substituting into the difference equation for n ≥ 0

            K0 + K1*n –a[(K0 – K1) +K1*n = 1

            Or

            [K0*(1-a) + a* K1] + [K1(1-a)]*n = 1

            Therefore

            K1 = 0 and K0 = 1/(1-a)

            The total solution is the sum of the homogeneous and particular solutions so

            yn = KH*an +1/(1-a)

            but y0 = 1 so

            1 = KH*a0 +1/(1-a)

            Or KH = 1 – 1/(1-a) = -a/(1-a)

            This makes the total solution

            yn = -a(n+1)/(1-a) + 1/(1-a)

            Or

            yn = [1-a(n+1)]/(1-a) for n ≥ 0

Which agrees with our previous solution.