# Properties of Discrete Time Unit Impulse Signal

Electronics & ElectricalElectronDigital Electronics

## What is a Discrete Time Impulse Sequence?

The discrete time unit impulse sequence 𝛿[𝑛], also called the unit sample sequence, is defined as,

$$\mathrm{\delta \left [ n \right ]=\left\{\begin{matrix} 1\; for\: n=0\ 0\; for \: n\neq 0\ \end{matrix}\right.}$$

## Properties of Discrete Time Unit Impulse Sequence

• ### Scaling Property

According to the scaling property of discrete time unit impulse sequence,

𝛿[𝑘𝑛] = 𝛿[𝑛]

Where, k is an integer.

Proof − By the definition of the discrete time unit impulse sequence,

$$\mathrm{\delta \left [ n \right ]=\left\{\begin{matrix} 1\; for\: n=0\ 0\; for \: n\neq 0\ \end{matrix}\right.}$$

Similarly, for the scaled unit impulse sequence,

$$\mathrm{\delta \left [ kn \right ]=\left\{\begin{matrix} 1\; for\: kn=0\ 0\; for \: kn\neq 0\ \end{matrix}\right.}$$ $$\mathrm{\Rightarrow \delta \left [ kn \right ]=\left\{\begin{matrix} 1\; for\: n=\frac{0}{k}=0\ 0\; for \: n\neq\frac{0}{k}\neq 0\ \end{matrix}\right.=\left\{\begin{matrix} 1\; \; for\: n=0\ 0\; \; for\: n\neq 0\ \end{matrix}\right.=\delta \left [ n \right ]}$$
• ### Product Property

𝑥[𝑛]𝛿[𝑛 − 𝑛0] = 𝑥[𝑛0]𝛿[𝑛 − 𝑛0]

Proof − By the definition of the unit impulse signal, we know,

$$\mathrm{\delta \left [ n-n_{0} \right ]=\left\{\begin{matrix} 1\: \: for\: n=n_{0}\ 0\: \: for\: n\neq n_{0}\ \end{matrix}\right.}$$

As from the expression it is clear that the impulse sequence has a non-zero value only at 𝑛 = 𝑛0. Therefore,

𝑥[𝑛]𝛿[𝑛 − 𝑛0] = 𝑥[𝑛0]𝛿[𝑛 − 𝑛0]

• ### Shifting Property

$$\mathrm{x\left [ n \right ]=\sum_{k=-\infty }^{\infty }x\left [ k \right ]\delta \left [ n-k \right ]}$$

Proof − By using the product property of the discrete time unit impulse sequence, we have,

𝑥[𝑛]𝛿[𝑛 − 𝑛0] = 𝑥[𝑛0]𝛿[𝑛 − 𝑛0]   … (1)

Putting k in place of 𝑛0 in equation (1), we get,

𝑥[𝑛]𝛿[𝑛 − 𝑘] = 𝑥[𝑘]𝛿[𝑛 − 𝑘]

$$\mathrm{\Rightarrow \sum_{k=-\infty }^{\infty }x\left [ n \right ]\delta \left [ n-k \right ]=\sum_{k=-\infty }^{\infty }x\left [ k \right ]\delta \left [ n-k \right ]}$$ $$\mathrm{\Rightarrow x\left [ n \right ]\sum_{k=-\infty }^{\infty }\delta \left [ n-k \right ]=\sum_{k=-\infty }^{\infty }x\left [ k \right ]\delta \left [ n-k \right ]}$$ $$\mathrm{\because \sum_{k=-\infty }^{\infty }\delta \left [ n-k \right ]=1}$$ $$\mathrm{\therefore x\left [ n \right ]=\sum_{k=-\infty }^{\infty }x\left [ k \right ]\delta \left [ n-k \right ]}$$
• ### The discrete time unit impulse sequence is the first difference of discrete time unit step sequence. That is,

𝛿[𝑛] = 𝑢[𝑛] − 𝑢[𝑛 − 1]

Proof − By the definition of the discrete time unit step sequence,

$$\mathrm{u\left [ n \right ]=\sum_{k=0}^{\infty }\delta \left [ n-k \right ] =\delta \left [ n \right ]+\sum_{k=1}^{\infty }\delta \left [ n-k \right ] }$$ $$\mathrm{\because u\left [ n-1 \right ]=\sum_{k=1}^{\infty }\delta \left [ n-k \right ]}$$

∴ 𝑢[𝑛] = 𝛿[𝑛] + 𝑢[𝑛 − 1]

⟹ 𝛿[𝑛] = 𝑢[𝑛] − 𝑢[𝑛 − 1]