### Required InformationConsider the signal \( x(t) \) where the harmonic function is based on \( T = T_0 \), which is the fundamental period of \( x(t) \).\( x(t) = -2\sin(4\pi t) + 5\cos(12\pi t) \)Consider \( c_x[k] \) to be the harmonic function of \( x(t) \).\[ c_x[k] = jA \left( \delta[k+a] - \delta[k-a] \right) + B \left( \delta[k-b] + \delta[k+b] \right) \]The numerical values of the constants are \( A = \_\_\_\_\_\_ \), \( a = \_\_\_\_\_\_ \), \( B = \_\_\_\_\_\_ \), and \( b = \_\_\_\_\_\_ \).

Given: Brief description: In the given question we have a continuous time domain signal called x(t). First of all we need to calculate the value of variables which are mentioned in the question. By writing the Fourier series coefficients of the given signal we can calculate the variables of the circuit. Ck = Exponential Fourier series coefficient By writing the given continuous time domain signal into its exponential and comparing with the standard Fourier series expression we have calculated all the variable.Note : exp = Exponential function x(t) = -2 sin( 4πt ) +5 cos ( 12πt) The time period of the given signal is calculated as Wm = m linear frequency of the given time domain signal W1 = 4π rad/sec W2 = 12π rad/sec Wo = fundamental frequency of a given time domain signal Wo = GCD( W1 ,W2 ) Wo = GCD( 4π , 12π) Wo = 4π rad/sec Wo = 2π/ To To = (2π) / Wo To = ( 2π ) / 4π To = 1/2 sec To = 0.5 sec The standard Fourier series of a continuous time domain signal is given by x(t) = Σ Ck exp(jkwot) -------(1) Cn =Exponential Fourier series coefficients of the given time domain signal wo = fundamental frequency of the time domain signal. x(t) = -2 sin( 4πt ) +5 cos ( 12πt) Writing the given signal in the form of exponential we get as, x(t) = (-2) [ exp(j4πt) - exp(-j4πt) ]/ (2j) + 5 (exp(j12πt) + exp(-j12πt) ) / (2) Further simplifying the above expressions we get x(t) = [ j exp(j4πt) - j exp(j4πt) + 2.5 exp(j12πt) + 2.5 exp(-j12πt) ] x(t) = [ ( j exp(j×1× Wot) - j exp(j×1× Wot) ) + (2.5 exp(j3Wot) + 2.5 exp(-j3Wot) ) ] -----(2) The exponential Fourier series coefficients are, C1 = j C-1 = -j C3 = 2.5 C-3 = 2.5 Writing the given function into impulse function form we get as Ck = j δ( k -1) - j δ( k + 1) + 2.5 δ( k -3) + 2.5 δ( k + 3) Rewriting the given expression in the form of question Ck = (-j ) [ δ( k + 1) - δ( k -1) ] + 2.5 [ δ( k -3) +δ( k + 3)] By comparing with the given standard question the values of variables A = -1 a = 1 B = 2.5 b = 3