 Hello, welcome all. Welcome to video lecture on properties of Z-transform. Myself, Mr. S. N. Chamath Goudar from Walchand Institute of Technology, Solapur. At the end of this session, students will be able to explain the properties of Z-transform. The various properties of the Z-transform are as listed below. Linearity, time shifting, scaling in the Z-domain, time reversal, the convolution property and differentiation in the Z-domain. So, today first property we are going to discuss is linearity. This property states that if x 1 of n is a time domain discrete time signal, then Z-transform of x 1 of n is equal to x 1 of z with R o c is equal to R 1. Similarly, if x 2 of n is a discrete time signal, then Z-transform of x 2 of n is equal to x 2 of z with R o c is equal to R 2. Now, we scale the discrete time signals x 1 of n and x 2 of n by a and b. So, that a into x 1 of n is a discrete time signal and Z-transform of a into x 1 of n will be equal to a into x 1 of z with R o c is equal to R 1. Similarly, b is a scaling factor of x 2 of n and b into x 2 of n is a discrete time signal and Z-transform of b into x 2 of n is equal to b into x 2 of z with R o c is equal to R 1. Now, if a into x 1 of n plus b into x 2 of n is a discrete time signal, then Z-transform of a into x 1 of n plus b into x 2 of n will be equal to a into x 1 of z plus b into x 2 of z with R o c containing R 1 intersect R 2. Now, let us prove this one. We know that x of z is equal to summation n equal to minus infinity to infinity x of n z to the power minus n, where x of n is equal to a into x 1 of n plus b into x 2 of n. Now, by substituting x of n equal to a into x 1 of n plus b into x 2 of n in the above expression, we get x of z is equal to summation n equal to minus infinity to plus infinity a into x 1 of n plus b into x 2 of n into z to the power minus n. Now, the summation can be split into two parts n equal to minus infinity to plus infinity a into x 1 of n into z to the power minus n plus b summation n equal to minus infinity to plus infinity b into x 2 of n into z to the power minus n. So, in the above expression a and b are the constant terms. So, we will take them outside the summation here. So, x of z will be written as a into summation n equal to minus infinity to infinity x 1 of n z to the power minus n plus b into summation n equal to minus infinity to infinity x 2 of n z to the power minus n. So, we know that summation n equal to minus infinity to infinity x 1 of n into z to the power minus n is equal to x 1 of z and summation n equal to minus infinity to infinity x 2 of n into z to the power minus n is equal to x 2 of z. Hence, proved. Now, let us discuss time shifting property. Time shifting property states that if x of n is a discrete time signal, then z transform of x of n is equal to x of z with r o c is equal to r. Then x of n minus n naught is a is a discrete time signal, then z transform of x of n minus n naught is equal to z to the power minus n naught into x of z with r o c r o c is equal to r except for the possible addition or deletion of origin or infinity. Let us prove this one. We know that x of z is equal to summation n equal to minus infinity to infinity x of n z to the power minus n where x of n is equal to x of n minus n naught. Now, by substituting x of n equal to x of n minus n naught in the above expression x of z is equal to summation n equal to minus infinity to infinity x of n minus n naught z to the power minus n. Now, x of z is equal to summation n equal to minus infinity to infinity x of n minus n naught into z to the power minus n here. Let us put n minus n naught is equal to k. Then n is equal to k plus n naught minus n is equal to minus k minus n naught. If n is equal to minus infinity, then n minus n naught will be equal to minus infinity minus n naught. We replace n minus n naught by k here. So, k is equal to minus infinity minus subtracting minus n naught from minus infinity will not affect and it remains same that is minus infinity. If n is equal to plus infinity, then n minus n naught is equal to infinity minus n naught. We replace n minus n naught by k here k is equal to plus infinity. Substracting minus n naught from infinity will not affect the infinite value. So, it remains same that is infinity. Now, we substitute these values k is equal to plus infinity k is equal to minus infinity and minus n is equal to minus k minus n naught and n minus n naught is equal to k in the above expression. Then it is written as summation k is equal to minus infinity to infinity x of k z to the power minus k minus n naught. So, this will be written as summation k is equal to minus infinity to infinity x of k z to the power minus k into z to the power minus n naught. So, z to the power minus n naught is a constant term for this summation. So, we will take that one outside z to the power minus n naught into summation k is equal to minus infinity to infinity x of k z to the power minus k where summation k equal to minus infinity to infinity x of k z to the power minus k is equal to x of z. Therefore, this will be written as z to the power minus n naught into x of z. Hence, proved. These are the references. Thank you.