Solutions to Midterm Exam

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Lawrence Hoover
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1 Solutions to Midterm Exam. For the function: f [x] =sin[πx] RECT [x ] (a) Find expressions for the even and odd parts of f [x] (b) Sketch these even and odd parts. It will first be useful to sketch f [x]: notethat 0 if x 3 f [x] = sin [πx] if + <x< 3 0 if x and that THE PERIOD OF THE SINE IS UNIT, NOT (a mistake several people made). f[x] x The expressions for even and odd parts: f even [x] = (f [x]+f [ x]) f odd [x] = (f [x] f [ x]) From this we can sketch f [ x] = sin[π ( x)] RECT [ x ] = sin[ πx] RECT [ (x +)] = sin [πx] RECT [x +]

2 Sketch that: f[x] x f even [x] = (f [x]+f [ x]) = (sin [πx] RECT [x ] + ( sin [πx] RECT [x +])) = (sin [πx] RECT [x ] sin [πx] RECT [x +]) = sin [π x ] (RECT [x ] RECT [x +]) even part x

3 f odd [x] = (f [x] f [ x]) = (sin [πx] RECT [x ] ( sin [πx] RECT [x +])) = (sin [πx] RECT [x ] + sin [πx] RECT [x +]) = sin [πx] (RECT [x +]+RECT [x ]) odd part x 3

4 . For the function f [x] graphed below as real and imaginary parts (a) Find an expression for F [ξ] First we need an expression for f [x], by inspection, f [x] is a reversed and scaled replica of STEP [x] exp [ x] centered at x =, where the value of the scale factor b 0 is to be determined from the sketch. You may want to do this in stages sketchstep [x] exp [ x], then insert the scale factor b 0 to set the amplitude to be exp [ π] at x =+ STEP [x] exp [ πx] then scale the argument of the step function (has no impact on the amplitude: STEP [πx] exp [ πx] the reverse it by substituting x: x STEP [ πx] exp [ ( πx)] = STEP π then translate it by convolving with δ [x +]: x f [x] =STEP exp π x π exp δ [x +] Now apply the scaling and filter theorems to the known transforms: x π F{exp [ x] STEP [x]} = +πiξ = F{exp [ ( x)] STEP [ x]} = +πi ( ξ) = i πξ ½ = F exp x ¾ x STEP = π π π +πi ( π ξ) = π 4 µ i ξ

5 F{δ [x +]} = exp[ i πξ ( )] = exp[+i 4πξ] F [ξ] = F{0[x]+i (exp [+x] STEP [+x]) δ [x +]} = 0[ξ]+i exp [+i 4πξ] µ π i ξ = i π exp [+i 4πξ] ( + i ξ) ( i ξ) ( + i ξ) = i +i ξ exp [+i 4πξ] π +(ξ) 5

6 (b) sketch F [ξ] as magnitude and phase. First, evaluate the magnitude it is just the magnitude of the spectrum of the decaying exponential: F [ξ] = i +i ξ exp [+i 4πξ] π +(ξ) = i π +i ξ exp [+i 4πξ] +(ξ) = π +i ξ +(ξ) s µ = +i ξ π +(ξ) s +i ξ = π +(ξ) i ξ +(ξ) v = π u t +(ξ) +(ξ) µ +i ξ +(ξ) = π +(ξ) +(ξ) = π +(ξ) This is a relative of the Lorentzian function: F[xi] xi 6

7 Thephaseisthesumofthephasesofthethreecomponents: Φ {F [ξ]} = Φ {i} + Φ {exp [+i πξ]} + Φ = + π ξ +πξ +tan = + π +πξ +tan [ξ] ½ ¾ +i ξ +(ξ) Plot the inverse tangent first and then add the constant and the linear phase: Phase (units of pi) xi - -3 phase in units of π radians: red = tan [ξ] (from the decaying exponential), blue = ξ (from the linear phase), brown = + π (from the constant i) phase xi phase of entire spectrum (sum of the three terms just graphed) The phase is dominated by the contribution from the linear term 7

8 (c) (OPTIONAL BONUS), sketch the real and imaginary parts of F [ξ] F [ξ] = i π = i π exp [+i 4πξ] +i ξ +(ξ) (cos [4πξ]+i sin [4πξ]) +i ξ +(ξ) = +i ξ (i cos [4πξ] sin [4πξ]) π +(ξ) = ( sin [4πξ] ξ cos [4πξ]) π +(ξ) + i π (cos [4πξ] ξ sin [4πξ]) +(ξ) Real xi Real part of spectrum, which is the product of an odd sinusoid and a Lorentzian. Imaginary xi -0. Imaginary part of spectrum, which is even because of the factor of i in f [x] Note that the imaginary part is even and the real part is odd, because of the factor of i

9 3. For the three functions f [x] = SINC [x +] f [x] = SINC [x] f 3 [x] = SINC [x ] find an expression for g [x] such that: g [x] =f [x] f [x] f 3 [x] Solution: first, I suggest expression the translated SINC functions as convolutions: SINC [x x 0 ] = SINC [x] δ [x x 0 ] = SINC [x ± ] = SINC [x] δ [x ± ] g [x] = f [x] f [x] f 3 [x] =SINC [x] δ [x +] SINC [x] SINC [x] δ [x ] = SINC [x] SINC [x] SINC [x] δ [x +] δ [x ] so we have the convolution of five functions, rather than three. Can now apply the filter theorem: G [ξ] = F {SINC [x]} F {SINC [x]} F {SINC [x]} F{δ[x +]} F{δ [x ]} = RECT [ξ] RECT [ξ] RECT [ξ] exp [ πiξ ( )] exp [ πiξ (+)] = RECT [ξ] RECT [ξ] RECT [ξ] = RECT [ξ] g [x] =F {RECT [ξ]} = SINC [x] 9

10 4. If the -D Fourier transform of f [x] is F {f [x]} = F [ξ], derive the expression for G [ξ] that satifies F {G [ξ]} = i f [ x]. In other words, what is: F [f [ x]] = F ½f F [f [x]] = F { i f [ x]} = i F {f [ x]} by linearity = = Z + Z + µz + ¾ x = F [ ξ] =F [ ξ] by scaling theorem f [x] e πiξx dx (f [x] exp[+πiξx]) dx f [x] exp[ πi ( ξ) x] dx = F [ ξ] F [f [x]] = F [ ξ] = F{f [ x]} = F [+ξ] = F{ i f [ x]} = i F [+ξ] = G [ξ] =F { i f [ x]} = i F [+ξ] G [ξ] = i F [+ξ] 0

11 5. Evaluate the autocorrelation of f [x] =exp ³ x h x i iπ RECT Solution: f [x] Ff [x] = g [x] F{f [x] Ff [x]} = F{g [x]} = F [ξ] = g [x] =F F [ξ] ª ½ ³ x h x i ¾ F [ξ] = F exp iπ RECT µ ½ ³ x ¾ ³ n h x io = F exp iπ F RECT ³ h = exp i π i exp +iπ (ξ) ( SINC [ξ]) h 4 = 4 exp i π i exp +iπ (ξ) SINC [ξ] ³ h 4 F [ξ] = 4 exp i π i exp +iπ (ξ) SINC [ξ] ³ 4 h 4 exp i π i exp +iπ (ξ) SINC [ξ] 4 h = 4 exp i π i exp +iπ (ξ) SINC [ξ] 4 = 6 SINC [ξ] = 6 SINC [ξ] g [x] = F F [ξ] ª =6 h x i g [x] =8 TRI µ h x i h x i TRI =8 TRI

12 6. Evaluate the area of f [x] =exp[ x ] We would like to use the central ordinate theorem, so we need to find F [ξ], whichis real-valued and even. From the graph: f[x] x we can see that this is the sum of a scaled decaying exponential and its reversed replica: f [x] =exp[ x] STEP [x]+exp[+x] STEP [ x] The spectrum of the scaled decaying exponential is easily obtained from the scaling theorem: r [x] = exp xb0 STEP [x] =exp xb0 x STEP R [ξ] = b 0 +i πb 0 ξ F{exp [ x] STEP [x]} = +i π ξ = +iπξ = F{exp [+x] STEP [ x]} = = iπξ +iπ ( ξ) b 0

13 Central ordinate theorem: F [ξ] = +iπξ + iπξ = ( iπξ)+(+iπξ) ( + iπξ) ( iπξ) = +(πξ) = +(πξ) Z + f [x] dx = F [0] = +(πξ) Z + ξ=0 exp [ x ] dx = = 3

14 7. Describe in words and make a sketch that shows the significant features of δ [cos [πx ]]; I m not looking for exact expressions or sketches, just evidence of understanding of the relevant principles. Recall the formula for the Direac delta function if the argument is itself a function that hasnzerosatthelocationsx n δ [g [x]] = NX δ [x x n ] g x=xn n= The scaling of the area of each Dirac delta function is the reciprocal of the slope of g [x] at each zero, so steeper slopes lead to smaller weights applied to the Dirac delta function.. The functional argument of the Dirac delta function is the chirp function cos [πx ], which we have seen before. Clearly the absolute values of the slopes at the zero crossings increase with increasing distance from the origin, which means that the weights applied to the Dirac delta functions decrease with increasing distance. Thezerosarelocatedattheargumentswherecos [πx ]=0: πx = ± π, 3π µ n +, = ± π r = x = ± n + where n =0,,, 3, The slopes are: d cos πx = d cos (u) where u = πx dx dx = d du cos [u] du dx = sin πx d πx dx = sin πx πx d cos πx = π x sin πx dx which is evaluated at the zeros. Since cos [πx ]=cos π ( x) (i.e., the cosine chirp is an EVEN function), then we get the same zeros on both sides of the origin. For the dx 4

15 positivevaluesofx,weget: δ cos πx = = n=0 n=0 = π h x i n + NX δ for x>0 π x sin [πx ] h i NX δ x n + π n + sin π n + h i h i 3 δ x δ x h δ x 5 5 i + ' (.44 δ [x 0.707] δ [x.5] δ [x.58] + ) π For the negative values of x, wegetthesameweights: δ cos πx = π h δ i x + + h δ x i + h δ x i + for x<0 ³ Amplitudes (i.e., areas) of Dirac delta functions decrease with increasing x as n+ 5

16 8. Consider an imaging system with impulse response h [x] and input f [x] as listed: h [x] = exp +iπx f [x] = δ [x + x 0 ]+δ[x x 0 ] that produces the output g [x]: g [x] = f [x] h [x] (a) Find the general expression for g [x]; your result should include x 0 as a parameter. You may evaluate this as a convolution or by applying the filter theorem; I ll do the convolution first: g [x] = (δ [x + x 0 ]+δ [x x 0 ]) exp +iπx (δ [x + x 0 ]+δ[x x 0 ]) exp +iπx = δ [x + x 0 ] exp +iπx + δ [x x 0 ] exp +iπx = exp +iπ (x + x 0 ) +exp +iπ (x x 0 ) = exp +iπ x + x 0 +xx 0 +exp +iπ x + x 0 xx 0 = exp +iπ x + x0 (exp [+iπxx0 ]+exp[ iπxx 0 ]) = exp +iπ x + x0 cos[+π (xx0 )] This is a cosine with period x 0. Now evaluate the suared magnitude: g [x] = exp +iπ x + x 0 cos[+πxx0 ] = exp +iπ x + x 0 cos[+πxx 0 ] = cos[+πxx 0 ] = 4 cos [πx 0 x] g [x] =4 cos [πx 0 x] Note that it is useful (though not essential) to apply the identity for cos [θ] cos [θ] = + cos [θ] = g [x] =4 cos [πξx 0 ]=+cos[π (x 0 ) x] g [x] =+cos[π (x 0 ) x] This is a biased cosine with period (x 0 )! 6

17 (b) Sketch g [x] for two values of x 0 : x 0 =unit and x 0 =units. g [x] = 4 cos [π x] g [x] = 4 cos [π x] g[x] g [x] =4 cos [πx] in red and g [x] =4 cos [π x] in blue x (c) Describe the difference in the output as x 0 is increased from x 0 =0. The oscillation rate of g [x] INCREASES with increasing x 0 ; you have just evaluated the model for interference of two point sources in the Fresnel diffraction region. 7

Comments Histogram of Scores (maximum = 90 without extra credit) Statistics of Scores Problem a b a b c 3 4 5 6 7 8a 8b 8c 8 Total Score 7 8 5 9 6 5 5 5 5 5 5 7 6 5 90 μ 6.3 6.5.38 3.53.3 4.38.6 9.

18 Comments Histogram of Scores (maximum = 90 without extra credit) Statistics of Scores Problem a b a b c a 8b 8c 8 Total Score μ σ Max Min Med From the statistics, the problems that caused the greatest difficulty are obvious # (the decaying exponential), #4 (find G [ξ] such that F {G [ξ]} = i f [ x]), and #8 (the imaging system with output g [x] = f [x] h [x]. These difficulties appear to be due to (a) people not remembering the definitions functions and (b) people not remembering the theorems. 8

Problem μ Max Min

Problem μ Max Min SIMG-733 Optics-9 Solutions to Midterm Eam Statistics and Comments: Man people are still not sketching before writing down equations. Some wrote down equations with no eplanation, e.g., the depth of field