Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The FS coefficients of a signal $x(t)$ is given by

$$x(t) \longrightarrow C_k$$

The frequency shift property says that: if we multiply a signal $x(t)$ by $e^{j m\omega_0 t}$ the fourier series coefficients are now defined by $C_{k-m}$.

$$e^{j m \omega_0 t}x(t) \longrightarrow C_{k-m}$$

Now consider a signal $x(t) = \sin \dfrac{\pi}{4}t$

The fourier series coefficients are $C_1=\dfrac{1}{2j},C_{-1}=\dfrac{-1}{2j}$.

Now consider a signal $y(t) = e^{j\frac{\pi}{4} t}x(t)=e^{j\frac{\pi}{4} t}\sin \dfrac{\pi}{4} t$

The fourier series coefficients according to the property should be $(m=1)$:

$$C_{1-1}=C_0=\dfrac{1}{2j}$$ $$C_{-1-1}=C_{-2}=-\dfrac{1}{2j}$$

But if we expand the $e^{j\frac{\pi}{4} t}\sin \dfrac{\pi}{4} t$ using the formula for $\sin t = \dfrac{e^{jt}-e^{-jt}}{2j}$ we get the FS Coefficients as

$$C_0 = -\dfrac{1}{2j}$$ $$C_2 = \dfrac{1}{2j}$$

Can somebody explain this behavior? Am I using the property correctly or not ?


share|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.