What is the time period of $$ x(t) = 7 e^{\jmath(5t + \pi/2)} + 10^{\jmath(7t + \pi/5)}$$ ?

X(t) is the combination of two functions. one is the natural logarithm base, $e$, and the other is 10.

if there is only an exponential function given then we easily calculate the period of the given signal.

but the problem is another function given as $ 10^{j(7t+\pi/5)} $

How is the periodicity calculated if $e$, the natural logarithm base, is not in the base?

edit yes, it was asked in competitive exams, not as homework.

if $$ 7 e^{\jmath(5t + \pi/2)} $$

then perodicity we calculate as $$2\pi/w$$ so perodicity of exponetial part is $$2\pi/5 $$

i have confusion in this part $ 10^{j7t+\pi/5} $ is perodicity we calculate like same as above $$2\pi/w = 2\pi/7$$ is this correct?

  • $\begingroup$ Cool question! Just a small concern about MathJax; the \jmath renders as a "J" with no dot on top (in your original equation $$ x(t) = 7 e^{\jmath(5t + \pi/2)} + 10^{\jmath(7t + \pi/5)}$$ but later you use a bare "j" and it looks like what I'm used to for the representation of the square root of negative one: $$ 10^{j7t+\pi/5} $$ with a dot over the lower-case "j". Is the mixed notation intentional and meaningful? (zoomed screenshot i.sstatic.net/52pgW.png) $\endgroup$
    – uhoh
    Jan 8, 2023 at 3:05

1 Answer 1


A signal $e^{j(\omega t + \varphi)}$ has an angular frequency $\omega$ and period $T=2\pi/\omega$.

The signal $x_1(t) = 7e^{j(5t + \pi/2)}$ thus has period $$T_1 = \frac{2\pi}{5}$$

Since $a^n = e^{n \cdot \ln(a)}$, the signal $x_2$ can be rewritten as $x_2(t) = 10^{j(7t + \pi/5)} = e^{j(7 \cdot \ln(10) \cdot t + \ln(10) \cdot \pi/5)}$ and therefore has a period $$T_2 = \frac{2\pi}{7 \cdot \ln(10)}$$

Since $\ln(10)$ is irrational, $T_1 / T_2 = 7/5 \cdot \ln(10)$ is irrational too, and therefore the signal $x(t) = x_1(t) + x_2(t)$ is not periodic.

Most probably, it might simply be a typo in the question of the exam, which was supposed to be $$x(t) = 7e^{j(5t + \pi/2)} + 10e^{j(7t + \pi/5)}$$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.