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The sum of two continuous time signals 𝑥_{1}(𝑡) and 𝑥_{2}(𝑡) can be obtained by adding their values at every instant of time. Likewise, the difference of two continuous time signals 𝑥_{1}(𝑡) and 𝑥_{2}(𝑡) can be obtained by subtracting the values of one signal (say 𝑥_{2}(𝑡)) from another signal (say 𝑥_{1}(𝑡)) at every instant of time.

Let two continuous time signals 𝑥_{1}(𝑡) and 𝑥_{2}(𝑡) as shown in Figure-1. The sum of these two signals 𝑥_{1}(𝑡) + 𝑥_{2}(𝑡) is also shown in Figure-1.

The sum of these two signals can be obtained by considering different time intervals as follows −

**For 𝟎 ≤ 𝒕 ≤ 𝟏**− 𝑥_{1}(𝑡) = 3 and 𝑥_{2}(𝑡) is increasing linearly from 0 to 2. Therefore, the sum [i.e., 𝑥_{1}(𝑡) + 𝑥_{2}(𝑡)] will also increase linearly from (3 + 0 = 3) to (3 + 2 = 5).**For 𝟏 ≤ 𝒕 ≤ 𝟑**− 𝑥_{1}(𝑡) = 2 and 𝑥_{2}(𝑡) = 2, thus𝑥

_{1}(𝑡) +𝑥_{2}(𝑡) = 2 + 2 = 4**For 3≤ 𝒕 ≤ 𝟒**− 𝑥_{1}(𝑡) = 3 and 𝑥_{2}(𝑡) decreases linearly from 2 to 0. Hence, the sum of the signals will also fall linearly from (3 + 2 = 5) to (3 + 0 = 3).

The subtraction [𝑥_{1}(𝑡) − 𝑥_{2}(𝑡)] of two continuous time signals 𝑥_{1}(𝑡) and 𝑥_{2}(𝑡) is shown in Figure-2.

The difference of two continuous-time signals 𝑥_{1}(𝑡) and 𝑥_{2}(𝑡) can be obtained by considering different time intervals as follows −

**For 𝟎 ≤ 𝒕 ≤ 𝟏**−𝑥_{1}(𝑡) = 3 and 𝑥_{2}(𝑡) is increasing linearly from 0 to 2. Thus, the difference [𝑥_{1}(𝑡) −𝑥_{2}(𝑡)] falls linearly from (3 – 0 = 3) to (3 – 2 = 1).**For 𝟏 ≤ 𝒕 ≤ 𝟑**− 𝑥_{1}(𝑡) = 2 and 𝑥_{2}(𝑡) = 2, thus𝑥

_{1}(𝑡) − 𝑥_{2}(𝑡) = 2 − 2 = 0**For 3≤ 𝒕 ≤ 𝟒**− 𝑥_{1}(𝑡) = 3 and 𝑥_{2}(𝑡) decreases linearly from 2 to 0. Hence, the difference of the signals will rise linearly from (3 – 2 = 1) to (3 – 0 = 3).

In the discrete time case, the sum of two sequences 𝑥_{1}(𝑛) and 𝑥_{2}(𝑛) can be obtained by adding the corresponding sample values. Similarly, the difference of the two sequences 𝑥_{1}(𝑛) and 𝑥_{2}(𝑛) can be obtained by subtracting each sample of one signal from the corresponding sample of the other signal.

Let two discrete-time sequences are given as follows −

𝑥_{1}(𝑛) = {2, 1, 3, 5, 2}

𝑥_{2}(𝑛) = {1, 4, 2, 1, −3}

Then, **the addition of discrete-time signals** is,

𝑥_{1}(𝑛) + 𝑥_{2}(𝑛) = {2 + 1, 1 + 4, 3 + 2, 5 + 1, 2 − 3} = {3, 5, 5, 6, −1}

Similarly, **the subtraction of discrete-time signals** is,

𝑥_{1}(𝑛) - 𝑥_{2}(𝑛) = {2 − 1, 1 − 4, 3 − 2, 5 − 1, 2 + 3} = {1, −3, 1, 4, 5}

- Related Questions & Answers
- Signals and Systems: Multiplication of Signals
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- Signals and Systems – Classification of Signals
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