Discrete Time Systems Difference Equations - Free MCQ Practice Test

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

Explanation: 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 ‘Forced response’.

Explanation: 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 if such system is also known as Natural or Free response.

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

Explanation: The assumed solution obtained by assigning x(n)=0 is

Explanation: 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 – 3λ^n-1 – 4λ^n-2 = 0
=> λ^n-2(λ² – 3λ – 4) = 0
the roots of the above equation are λ=-1,4
Therefore, general form of the solution of the homogenous equation is

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 Y_zi(n) = (-1)ⁿ⁺¹ + (4)ⁿ⁺².

Explanation: The assumed solution of the difference equation to the forcing equation x(n), called the particular solution of the difference equation is
y_p(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=1/(1+a)
Therefore the particular solution is y_p(n)= 1/(1+a) u(n).

Explanation: The assumed solution of the difference equation to the forcing equation x(n), called the particular solution of the difference equation is
y_p(n)=Kx(n)=K2ⁿu(n) (where K is a scale factor)
Upon substituting y_p(n) into the difference equation, we obtain
K2ⁿu(n)=5/6K2^n-1u(n-1)-1/6 K2^n-2u(n-2)+2ⁿu(n)
To determine K we must evaluate the above equation for any n>=2, so that no term vanishes.
=> 4K= 5/6(2K)-1/6 (K)+4
=> K= 8/5
=> y_p(n)= (8/5) 2ⁿ.

Explanation: The linearity property of the linear constant coefficient difference equation allows us to add the homogenous and particular solution in order to obtain the total solution.

Explanation: The homogenous solution of the given equation is y_h(n)=C1(-1)ⁿ+C2(4)ⁿ—-(1)
To find the impulse response, x(n)=δ(n)
now, for n=0 and n=1 we get
y(0)=1 and
y(1)=3+2=5
From equation (1) we get
y(0)=C1+C2 and
y(1)=-C1+4C2
On solving the above two set of equations we get
C1=- 1/5 and C2= 6/5
=>h(n)= [-1/5 (-1)ⁿ + 6/5 (4)ⁿ]u(n).