Principles of Modern Radar - Volume 2 1891121537

54 CHAPTER 2 Advanced Pulse Compression Waveform ModulationsFIGURE 2-17 Thewaveform’snonlinear frequencyresponse willgenerate a Taylorweighted spectrum.frequency (normalized by β)0.50.40.30.20.10−0.1−0.2−0.3−0.4−0.5−0.5 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 0.5time (normalized by τ)At this point, we want to implement the graphical technique proposed by Cook [26].First, the group delay in equation (2.101) is evaluated at equally spaced points in .Next, a numerical interpolation is performed on the samples to obtain equally spacedpoints in time (i.e., group delay) that map to specific values of the independent frequencyvariable, . The time samples are then multiplied by –1 (flipping them about the groupdelay axis) and circularly rotated left to right (e.g., a clockwise rotation of a row vectorcontaining the samples) creating new function with time as the new independent variableand instantaneous frequency as the new dependent variable. Figure 2-17 contains a plotof the instantaneous frequency obtained from equation (2.101) after applying a cubicinterpolation and the inversion process.A model for the instantaneous frequency [3] is proposed:[t Mφ ′ (t) = 2πβτ + 2 ∑( ) ] 2πmtd k sin(2.102)τm=1Equation (2.102) asserts that the instantaneous frequency may be modeled as a linear termplus a sum of harmonically related and weighted sine functions. To obtain the coefficientsd k , the linear component is first subtracted from both sides of the equationφ ′ (t)2πβ − t M τ = 2 ∑( ) 2πmtd k sinτm=1(2.103)The right side of equation (2.103) represents the first M terms of the Fourier series ofan odd signal. Figure 2-18 contains a plot of the instantaneous frequency with the linearcomponent removed. The response is odd and may be interpreted as one period of aperiodic function. Harmonic analysis is applied to the signal in Figure 2-18 to obtain thecoefficients d k .A periodic signal x(t) may be expressed as a sum of weighted of sines and cosineswhere∞∑x(t) = a 0 + 2 [b k cos (m 0 t) + d k sin (m 0 t)] (2.104)m=1