Question

How is the discrete time impulse function defined in terms of the step function?

• A. \(\delta[n] = u[n+1] - u[n]\).
• B. \(\delta[n] = u[n] - u[n-2]\).
• C. \(\delta[n] = u[n] - u[n-1]\).
• D. \(\delta[n] = u[n+1] - u[n-1]\).

Hint:

Discrete Impulse and Step. The unit impulse \(\delta[n]\) is the first difference of the unit step \(u[n]\). \(\delta[n] = u[n] - u[n-1]\). Conversely, the unit step is the running sum of the unit impulse.

Answer 1

The Correct Option is C

The discrete-time unit impulse function, \(\delta[n]\), is defined as: $$ \delta[n] = \begin{cases} 1 & \text{if } n = 0
0 & \text{if } n \neq 0 \end{cases} $$ The discrete-time unit step function, \(u[n]\), is defined as: $$ u[n] = \begin{cases} 1 & \text{if } n \ge 0
0 & \text{if } n<0 \end{cases} $$ Let's evaluate the expression \(u[n] - u[n-1]\): - \(u[n-1]\) is the unit step function shifted right by 1 sample
$$ u[n-1] = \begin{cases} 1 & \text{if } n-1 \ge 0 \implies n \ge 1
0 & \text{if } n-1<0 \implies n<1 \end{cases} $$ - Now consider the difference \(u[n] - u[n-1]\): - For \(n<0\): \(u[n]=0\), \(u[n-1]=0\)
Difference = 0
- For \(n = 0\): \(u[0]=1\), \(u[0-1]=u[-1]=0\)
Difference = 1 - 0 = (1) - For \(n \ge 1\): \(u[n]=1\), \(u[n-1]=1\)
Difference = 1 - 1 = 0
The difference \(u[n] - u[n-1]\) is 1 only at n=0 and 0 otherwise, which is exactly the definition of \(\delta[n]\)
Therefore, \(\delta[n] = u[n] - u[n-1]\)