# Fourier series coefficent of $Y(2t+1)$?

Suppose the fourier series coefficient of Y(t) is Cn then what will be the fourier series coefficint of Y(2t+1)?

My doubt regarding this is if we do first shifting and then scaling ans will be $e^{jnw}Cn$ but if we do first scaling and the shifting then it will be $e^{0.5jnw)}Cn$

why is my answer changing and what is the correct procedure?

Here both operation will give the same graph ## 1 Answer

HINT: $$x(t) \Leftrightarrow c_{n}$$ $$x(at) \Leftrightarrow c_{n}$$ On scaling operation Only the frequency will be scaled $a$ times i.e $\omega' \mapsto w.a$ $$x(t \pm t_{0}) \Leftrightarrow c_{n}\cdot e^{\pm j \, k\, \omega_{0}\, t_{0}}$$ operation wise:For $A.y(at \pm t_{0})$ $$y(t) \rightarrow A.y(t) \underbrace{\rightarrow}_{shifting} y(t \pm t_{0}) \underbrace{\rightarrow}_{scaling} y(at \pm t_{0})$$

• why can't we do first scaling and then shifting in that case my result is changing its very difficult to write what i want to say – Rohit Jun 24 '17 at 6:39
• yes it will change because the plot of the signal will change , take any example like rect(t/8) then apply two different procedures , first scale then shift , then first shift then scale – Zeno San Jun 24 '17 at 6:44
• if we take x(t)=rect(t/8) and apply operation like x(2t+1) then it will not change if we do first scaling x(2{t+1/2}} and then shifting and other way first shifting like x(t+1) and then scaling x(2t+1) it will not change sir – Rohit Jun 24 '17 at 6:48
• take x(t)=rect(t/8) then do x(2t) then x(2t+1) this is our first operation, after that apply x(t+1) then x(2t+1) , what do you see ? – Zeno San Jun 24 '17 at 6:55
• Yes i will get the same graph in both operation sir i wil upload the image – Rohit Jun 24 '17 at 6:57