Discrete Time Difference Equation MCQ Quiz – Objective Question with Answer for Discrete Time System Difference Equation

The memory of the system, describes, in some cases, the ‘state’ of the system, the output of the system is called as ‘zero-state response’.

The zero-state response depends on the nature of the system and the input signal. Since this output is a response forced upon it by the input signal, it is also known as a ‘Forced response’.

For a zero-input response, the input is zero and the output of the system is independent of the input of the system. So, the response of such a system is also known as a Natural or Free response.

By making the input x(n)=0 we will get a homogeneous difference equation and the solution of that differential equation is known as a Homogenous or Complementary solution.

The assumed solution obtained by assigning x(n)=0 is
yh(n)=λn
=>y(n)+ay(n-1)=0
=>λn+a λn-1=0
=>λn-1(λ+A.=0
=>λ=-a
=>yh(n)=cλn=c(-A.n

Given difference equation is y(n)-3y(n-1)-4y(n-2)=0—-(1)
Let y(n)=λn

Substituting y(n) in the given equation
=> λn – 3λn-1 – 4λn-2 = 0
=> λn-2(λ2 – 3λ – 4) = 0

the roots of the above equation are λ=-1,4

Therefore, the general form of the solution of the homogenous equation is

yh(n)=C1 λ1n+C2 λ2n
=C1(-1)n+C2(4)n—-(2)

The zero-input response of the system can be calculated from the homogenous solution by evaluating the constants in the above equation, given the initial conditions y(-1) and y(-2).

From the given equation (1)
y(0)=3y(-1)+4y(-2)
y(1)=3y(0)+4y(-1)
=3[3y(-1)+4y(-2)]+4y(-1)
=13y(-1)+12y(-2)

From the equation (2)
y(0)=C1+C2 and
y(1)=C1(-1)+C2(4)=-C1+4C2

By equating these two set of relations, we have

C1+C2=3y(-1)+4y(-2)=15
-C1+4C2=13y(-1)+12y(-2)=65

On solving the above two equations we get C1=-1 and C2=16

Therefore the zero-input response is Yzi(n) = (-1)n+1 + (4)n+2.

The assumed solution of the difference equation to the forcing equation x(n), called the particular solution of the difference equation is
yp(n)=Kx(n)=Ku(n) (where K is a scale factor)

Substitute the above equation in the given equation
=>Ku(n)+aKu(n-1)=u(n)

To determine K we must evaluate the above equation for any n>=1, so that no term vanishes.

=> K+aK=1
=>K=\(\frac{1}{1+a}\)

Therefore the particular solution is yp(n)=\(\frac{1}{1+a}\) u(n).

The assumed solution of the difference equation to the forcing equation x(n), called the particular solution of the difference equation is
yp(n)=Kx(n)=K2nu(n) (where K is a scale factor)

Upon substituting yp(n) into the difference equation, we obtain

K2nu(n)=\(\frac{5}{6}\)K2n-1u(n-1)-\(\frac{1}{6}\) K2n-2u(n-2)+2nu(n)

To determine K we must evaluate the above equation for any n>=2, so that no term vanishes.

=> 4K=\(\frac{5}{6}\)(2K)-\(\frac{1}{6}\) (K)+4

=> K=\(\frac{8}{5}\)

=> yp(n)=(8/5) 2n.

The linearity property of the linear constant-coefficient difference equation allows us to add the homogeneous and particular solutions in order to obtain the total solution.