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# Fourier components of cos$\cos(2*pi*f1*t2\pi f_1t)$

I have the signal $$s(t) = cos(2\pi f_1t)$$$$s(t) = \cos(2\pi f_1t)$$ and I am looking for its components vs the Fourier basis, over the interval $$[0, T]$$. The formula for computing the coefficients is $$s_n = \int_{t_0}^{t_1} s(t) \frac{e^{-j\frac{2\pi}{T} nt}}{\sqrt{T}} dt$$$$s_n = \int_{t_0}^{t_1} s(t) \frac{e^{-j\frac{2\pi}{T} nt}}{\sqrt{T}} dt$$ and in this case $$t_0 = 0$$ and $$t_1 = T$$. The solution seems to be $$\frac{\sqrt{T}}{2}(-1)^n[Sinc(f_1T+n)e^{-j\pi f_1T} + Sinc(f_1T-n)e^{j\pi f_1T}]$$$$\frac{\sqrt{T}}{2}(-1)^n\bigg[\mathrm{sinc}\big(f_1T+n\big)e^{-j\pi f_1T} + \mathrm{sinc}\big(f_1T-n\big)e^{j\pi f_1T}\bigg]$$ but i can get no further than this $$\frac{\sqrt{T}}{2}\bigg(\frac{e^{j2\pi (f_1T-n)}-1}{j2\pi (Tf_1-n)} + \frac{e^{-j2\pi (f_1T+n)}-1}{-j2\pi (Tf_1+n)}\bigg)$$

# Fourier components of cos(2*pi*f1*t)

I have the signal $$s(t) = cos(2\pi f_1t)$$ and I am looking for its components vs the Fourier basis, over the interval $$[0, T]$$. The formula for computing the coefficients is $$s_n = \int_{t_0}^{t_1} s(t) \frac{e^{-j\frac{2\pi}{T} nt}}{\sqrt{T}} dt$$ and in this case $$t_0 = 0$$ and $$t_1 = T$$. The solution seems to be $$\frac{\sqrt{T}}{2}(-1)^n[Sinc(f_1T+n)e^{-j\pi f_1T} + Sinc(f_1T-n)e^{j\pi f_1T}]$$ but i can get no further than this $$\frac{\sqrt{T}}{2}\bigg(\frac{e^{j2\pi (f_1T-n)}-1}{j2\pi (Tf_1-n)} + \frac{e^{-j2\pi (f_1T+n)}-1}{-j2\pi (Tf_1+n)}\bigg)$$

# Fourier components of $\cos(2\pi f_1t)$

I have the signal $$s(t) = \cos(2\pi f_1t)$$ and I am looking for its components vs the Fourier basis, over the interval $$[0, T]$$. The formula for computing the coefficients is $$s_n = \int_{t_0}^{t_1} s(t) \frac{e^{-j\frac{2\pi}{T} nt}}{\sqrt{T}} dt$$ and in this case $$t_0 = 0$$ and $$t_1 = T$$. The solution seems to be $$\frac{\sqrt{T}}{2}(-1)^n\bigg[\mathrm{sinc}\big(f_1T+n\big)e^{-j\pi f_1T} + \mathrm{sinc}\big(f_1T-n\big)e^{j\pi f_1T}\bigg]$$ but i can get no further than this $$\frac{\sqrt{T}}{2}\bigg(\frac{e^{j2\pi (f_1T-n)}-1}{j2\pi (Tf_1-n)} + \frac{e^{-j2\pi (f_1T+n)}-1}{-j2\pi (Tf_1+n)}\bigg)$$

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# Fourier components of cos(2*pi*f1*t)

I have the signal $$s(t) = cos(2\pi f_1t)$$ and I am looking for its components vs the Fourier basis, over the interval $$[0, T]$$. The formula for computing the coefficients is $$s_n = \int_{t_0}^{t_1} s(t) \frac{e^{-j\frac{2\pi}{T} nt}}{\sqrt{T}} dt$$ and in this case $$t_0 = 0$$ and $$t_1 = T$$. The solution seems to be $$\frac{\sqrt{T}}{2}(-1)^n[Sinc(f_1T+n)e^{-j\pi f_1T} + Sinc(f_1T-n)e^{j\pi f_1T}]$$ but i can get no further than this $$\frac{\sqrt{T}}{2}\bigg(\frac{e^{j2\pi (f_1T-n)}-1}{j2\pi (Tf_1-n)} + \frac{e^{-j2\pi (f_1T+n)}-1}{-j2\pi (Tf_1+n)}\bigg)$$