1. Calculate the Fourier transform (X(jo)) of x(t): x(t) = cos(2nt)+cos(6nt); (2) Determine @OM, such that for arbitrary ||> @M, X(jo)=0; (3) Determine the minimal angular frequency, that satisfies Nyquist sampling theorem.
Please include solution on how to get the answers. Thank you
- The Fourier transform of x(t) can be calculated as follows:
X(jomega) = (1/2pi) * integral from -infinity to infinity of x(t)e^(-jomega*t) dt
X(jomega) = (1/2pi) * integral from -infinity to infinity of [cos(2nt)+cos(6nt)] * e^(-jomegat) dt
Using Euler's formula, we can rewrite cos(2nt) and cos(6nt) as complex exponential functions:
cos(2nt) = (e^(j2nt) + e^(-j2nt)) / 2 cos(6nt) = (e^(j6nt) + e^(-j6nt)) / 2
Substituting these expressions into the integral and simplifying, we get:
X(jomega) = (1/2pi) * [integral from -infinity to infinity of (1/2)e^(-j(2n-omega)*t) dt + integral from -infinity to infinity of (1/2)e^(-j(6n-omega)*t) dt]
Using the formula for the Fourier transform of a complex exponential function, we can evaluate each integral:
X(jomega) = (1/2pi) * [2pidelta(2n-omega)/2 + 2pidelta(6n-omega)/2]
Simplifying further, we get:
X(jomega) = (1/2)[delta(2n-omega) + delta(6n-omega)]
where delta is the Dirac delta function.
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