Note on fourier transform of unit step function

Note on fourier transform of unit step function

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  P a ge | 1 ADI DSP Learning Centre, IIT Madras A NOTE ON THE FOURIER TRANSFORM OF HEAVISIDE UNIT STEP FUNCTION S Anand Krishnamoorthy Project Associate, ADI DSP Learning Centre, IIT Madras I. INTRODUCTION The Heaviside unit step function is defined as follows – Table .I Continuous time Discrete time 𝒖(𝒕) = { 𝟏 ; 𝒕 ≥ 𝟎 𝟎 ; 𝒕 < 𝟎 } 𝒕 ∈ ℝ … (𝑰. 𝟏) 𝑢[𝑛] = { 1 ; 𝑛 ≥ 0 0 ; 𝑛 < 0 } 𝑛 ∈ ℤ … (𝐼. 2) Fig.1 Continuous and Discrete time Unit step signal The Unit step function can also be defined as the derivative of the standard ramp function- 𝑢(𝑡) = 𝑑 𝑑𝑥 max{𝑡, 0} … (𝐼. 3) Or as the integral of the Dirac delta function (impulse function) 𝑢(𝑡) = ∫ 𝛿(𝑠)𝑑𝑠 𝑡 −∞ … (𝐼. 4) The unit step function is causal, bounded in its amplitude but not stable, ∑ 𝑢[𝑛] ∞ −∞ = ∑ 1 ∞ 0 → ∞ … (𝐼. 5)
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  P a ge | 2 ADI DSP Learning Centre, IIT Madras The z transform of the unit step function is 𝐻[𝑧] = 1 1 − 𝑧−1 … (𝐼. 6) It is clear that the pole at 𝑧 = 1 lies on the unit circle and consequently, the Region of Convergence (ROC) is the region outside the unit circle. Fig.2 Z plane plot of Unit step signal The evaluation of the Fourier transform unit step function has been a confounding procedure. This is specifically due to its property that it is neither absolutely summable nor square summable. In the strict sense, the Fourier transform of the Heaviside unit step function does not exist. To incorporate the unit step function and other functions with similar properties in the Fourier domain, the Fourier transform must contain impulses at finite number of frequencies corresponding to the poles which lie on the unit circle. Thus, by allowing impulses in the frequency domain, it is possible to extend the Fourier transform representation to signals which are neither absolutely summable nor square summable. II. DERIVATION OF FOURIER TRANSFORM OF UNIT STEP FUNCTION II.1 Differential Approach The first approach for finding the Fourier transform of the unit step function is from the previous statement that the unit step function is the integral of the delta function. So, the derivation is as follows –
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  P a ge | 3 ADI DSP Learning Centre, IIT Madras We know that the Fourier transform of the impulse function is ℱ{𝛿(𝑡)} = 1 … (𝐼𝐼. 1) And 𝛿(𝑡) = 𝑑 𝑑𝑡 𝑢(𝑡) = 𝑢′(𝑡) … (𝐼𝐼. 2) And from the property of Fourier transform, ℱ{𝑋′(𝑡)} = 𝑗𝜔ℱ{𝑋(𝑡)} … (𝐼𝐼. 3) Therefore, ℱ{𝑢(𝑡)} = 𝑈(𝑗𝜔} = 1 𝑗𝜔 ℱ{𝛿(𝑡)} = 1 𝑗𝜔 … (𝐼𝐼. 4) But the above solution is not completely correct. Note that differentiation methods does not include the integral constants. Thus, the result must be strictly, ℱ{𝑢(𝑡)} = 1 𝑗𝜔 + ℱ{𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡} … (𝐼𝐼. 5) Fourier transform of a function is unique and has a one-to-one mapping. So, to find the constant term, the inverse Fourier transform of 1 𝑗𝜔 can be used- ℱ−1 { 1 𝑗𝜔 } = 1 𝑗2𝜋 ∫ 𝑒 𝑗𝜔𝑡 𝜔 𝑑𝜔 ∞ −∞ = 1 𝑗2𝜋 [∫ 𝑒 𝑗𝜔𝑡 𝜔 𝑑𝜔 0 −∞ + ∫ 𝑒 𝑗𝜔𝑡 𝜔 𝑑𝜔 ∞ 0 ] … (𝐼𝐼. 6) Combining into a single integral, we get ℱ−1 { 1 𝑗𝜔 } = 1 𝑗2𝜋 [− ∫ 𝑒−𝑗𝜔𝑡 𝜔 𝑑𝜔 ∞ 0 + ∫ 𝑒 𝑗𝜔𝑡 𝜔 𝑑𝜔 ∞ 0 ] = 1 𝑗2𝜋 ∫ 𝑒 𝑗𝜔𝑡 − 𝑒−𝑗𝜔𝑡 𝜔 𝑑𝜔 ∞ 0 … (𝐼𝐼. 7) Which can be simplified into ℱ−1 { 1 𝑗𝜔 } = 1 𝜋 ∫ sin 𝜔𝑡 𝜔 𝑑𝜔 ∞ 0 = 1 2 [∵ ∫ sin 𝜔𝑡 𝜔 𝑑𝜔 ∞ 0 = 𝜋 2 ] … (𝐼𝐼. 8) We can obtain − 1 2 if we evaluate in terms of the negative part of the integral. Thus, the inverse Fourier transform yields the function 𝑢1(𝑡) = { 1 2 ; 𝑡 ≥ 0 − 1 2 ; 𝑡 < 0 } … (𝐼𝐼. 9) But, from the equations I.1 and II.9, 𝑢(𝑡) = 𝑢1(𝑡) + 1 2 … (𝐼𝐼. 10) Thus, this is the term which was missing from the Fourier transform. Now the complete transform is given by- ℱ{𝑢(𝑡)} = 𝑈(𝑗𝜔} = 1 𝑗𝜔 + ℱ { 1 2 } = 1 𝑗𝜔 + 𝜋𝛿(𝜔) … (𝐼𝐼. 11)
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  P a ge | 4 ADI DSP Learning Centre, IIT Madras II.2 Signum function Approach Another approach followed by Oppenheim and others [2] is to derive the Fourier transform from another similar function – the signum function. This methods is very similar to the first method and yields the identical result. Consider the signum function 𝜎(𝑡) defined as 𝜎(𝑡) = { 1 ; 𝑡 ≥ 0 −1 ; 𝑡 < 0 } … (𝐼𝐼. 12) Fig.3 Continuous time Signum Function The Fourier transform of the signum function is given by ℱ{𝜎(𝑡)} = ∫ 𝜎(𝑡)𝑒−𝑗𝜔𝑡 𝑑𝑡 ∞ −∞ … (𝐼𝐼. 13) But we know that differential of the signum function is also the delta impulse function. Thus, 𝑑 𝑑𝑡 { 𝜎(𝑡) 2 } = 𝛿(𝑡) … (𝐼𝐼. 14) 𝑗𝜔 ℱ{𝜎(𝑡)} = 2ℱ{𝛿(𝑡)} … (𝐼𝐼. 15) ⇒ ℱ{𝜎(𝑡)} = 2 𝑗𝜔 [∵ ℱ{𝛿(𝑡)} = 1] … (𝐼𝐼. 16) The relationship between the unit step function and the signum function is given by 𝜎(𝑡) = 2𝑢(𝑡) − 1 … (𝐼𝐼. 17)
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  P a ge | 5 ADI DSP Learning Centre, IIT Madras Note the similarity of II.10 and II.17. ⇒ ℱ{𝜎(𝑡)} = 2ℱ{𝑢(𝑡)} − ℱ{1} … (𝐼𝐼. 18) ⇒ ℱ{𝑢(𝑡)} = 1 2 (ℱ{𝜎(𝑡)} + ℱ{1}) … (𝐼𝐼. 19) ⇒ ℱ{𝑢(𝑡)} = 1 2 ( 2 𝑗𝜔 + 2𝜋𝛿(𝜔)) … (𝐼𝐼. 20) ℱ{𝑢(𝑡)} = 𝑈(𝑗𝜔) = 1 𝑗𝜔 + 𝜋𝛿(𝜔) … (𝐼𝐼. 21) II.3 Z Transform Approach It is possible to derive the Fourier Transform of the unit step function is many ways. The second approach for finding the Fourier transform of the unit step function is as follows The Z transform of the unit step function is given by- 𝐻[𝑧] = ∑ 𝑢[𝑛]𝑧−𝑛 ∞ −∞ = ∑ 𝑧−𝑛 ∞ 0 = 1 1 − 𝑧−1 = 𝑧 𝑧 − 1 … (𝐼𝐼. 22) Substituting 𝑧 = 𝑒 𝑗𝜔 , except at 𝑧 = 1, 𝐻(𝑗𝜔) = 𝑒 𝑗𝜔 𝑒 𝑗𝜔 − 1 = 𝑒 𝑗𝜔 2 𝑒 𝑗𝜔 2 − 𝑒− 𝑗𝜔 2 = 𝑒 𝑗𝜔 2 2𝑗 sin 𝜔 2 … (𝐼𝐼. 23) This formula yields finite results at all frequencies except at 𝜔 = 2𝜋𝑘, 𝑘 ∈ {0,1,2, … } since 𝐻(𝑗𝜔) becomes indeterminate at those frequencies. This is due to the fact that the unit step signal is not piecewise continuous at the origin where there is an abrupt “jump”. Note that this approach disregards the impulse at the poles. II.4 Even Odd function Approach This is a very interesting approach, yet uses elementary concepts. This derivation, used in the Proakis et.al book [3] is as follows. Firstly, the discrete unit step function is written as a linear combination of an even and odd sequence. 𝑢[𝑛] = 𝑢 𝑒[𝑛] + 𝑢 𝑜[𝑛] … (𝐼𝐼. 24)
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  P a ge | 6 ADI DSP Learning Centre, IIT Madras Fig.4 Even and Odd parts of the Unit step signal Where the even function is 𝑢 𝑒[𝑛] = 1 2 [𝑢[𝑛] + 𝑢[−𝑛]] … (𝐼𝐼. 25) 𝑢 𝑒[𝑛] = 1 2 + 1 2 𝛿[𝑛] … (𝐼𝐼. 26) Taking Fourier transform of the even part, 𝑈𝑒(𝑗𝜔) = 𝜋𝛿(𝜔) + 1 2 … (𝐼𝐼. 27) And the odd part is given by 𝑢 𝑜[𝑛] = 1 2 [𝑢[𝑛] − 𝑢[−𝑛]] … (𝐼𝐼. 28) 𝑢 𝑜[𝑛] = 𝑢[𝑛] − 1 2 − 1 2 𝛿[𝑛] … (𝐼𝐼. 29) Now, shift the odd signal to the right by one sample, 𝑢 𝑜[𝑛 − 1] = 𝑢[𝑛 − 1] − 1 2 − 1 2 𝛿[𝑛 − 1] … (𝐼𝐼. 30)
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  P a ge | 7 ADI DSP Learning Centre, IIT Madras Subtracting II.31 from II.30, 𝑢 𝑜[𝑛] − 𝑢 𝑜[𝑛 − 1] = 𝑢[𝑛] − 𝑢[𝑛 − 1] − 1 2 [𝛿[𝑛] − 𝛿[𝑛 − 1]] … (𝐼𝐼. 31) ⇒ 𝑢 𝑜[𝑛] − 𝑢 𝑜[𝑛 − 1] = 𝛿[𝑛] − 1 2 [𝛿[𝑛] − 𝛿[𝑛 − 1]] = 1 2 [𝛿[𝑛] + 𝛿[𝑛 − 1]] … (𝐼𝐼. 32) Taking Fourier transform of the odd part and using the shifting property of the Fourier transform, 𝑈 𝑜(𝑗𝜔) [1 − 𝑒−𝑗𝜔 ] = 1 2 (1 + 𝑒−𝑗𝜔 ) … (𝐼𝐼. 33) 𝑈 𝑜(𝑗𝜔) = (1 + 𝑒−𝑗𝜔 ) 2(1 − 𝑒−𝑗𝜔) … (𝐼𝐼. 34) Now, we can combine the two individual Fourier transform to yield the final Fourier transform of the unit step function. ℱ{𝑢[𝑛]} = 1 2 (ℱ{𝑢 𝑒[𝑛]} + ℱ{𝑢0[𝑛]}) … (𝐼𝐼. 35) ℱ{𝑢[𝑛]} = 𝑈(𝑗𝜔) = 𝜋𝛿(𝜔) + 1 2 (1 + (1 + 𝑒−𝑗𝜔 ) (1 − 𝑒−𝑗𝜔) ) … (𝐼𝐼. 36) ⇒ 𝑈(𝑗𝜔) = 𝜋𝛿(𝜔) + ( 1 (1 − 𝑒−𝑗𝜔) ) … (𝐼𝐼. 37) III.CONCLUSION Now, there are three results for the Fourier transform of the Heaviside unit step signal, obtained from different methods. These results are tabulated in Table.II. Table. II Fourier transforms of Unit step signal Approach Fourier Transform 𝑼(𝒋𝝎) Differential approach & Oppenheim approach 𝜋𝛿(𝜔) + 1 𝑗𝜔 Z Transform approach 𝑒 𝑗𝜔 2 2𝑗 𝑠𝑖𝑛 𝜔 2 𝜔 ≠ 2𝜋𝑘, 𝑘 ∈ {0,1,2, … } Even Odd function approach (Proakis) 𝜋𝛿(𝜔) + ( 1 (1 − 𝑒−𝑗𝜔) )
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  P a ge | 8 ADI DSP Learning Centre, IIT Madras The result from the Z transform approach is not well-behaved at 𝜔 = 0. The other two results incorporate the impulse signal at the pole, thus can be considered valid. Note that the Oppenheim approach is for the continuous time unit step signal and the Proakis approach is for the discrete time unit step signal. This can be observed from their derivations. Therefore, both are valid in their own domain. Moreover, since the Z transform approach also deals in the discrete domain, it is equivalent to the even-odd approach except for the impulse at the pole. Observe that, 1 (1 − 𝑒−𝑗𝜔) = 1 𝑒−𝑗𝜔/2(𝑒 𝑗𝜔/2 − 𝑒−𝑗𝜔) = 𝑒−𝑗𝜔/2 2𝑗 𝑠𝑖𝑛 𝜔 2 , 𝜔 ≠ 2𝜋𝑘, 𝑘 ∈ {0,1,2, … } … (𝐼𝐼𝐼. 1) Therefore, Table.III Domain Fourier Transform 𝑼(𝒋𝝎) Continuous Unit step function 𝜋𝛿(𝜔) + 1 𝑗𝜔 Discrete Unit step function 𝜋𝛿(𝜔) + ( 1 (1 − 𝑒−𝑗𝜔) ) IV.REFERENCES [1] The Fourier Transform of the Heaviside function: A tragedy, Mathematical Physics Vol I, 2005 [2] A. Oppenheim, Shafer, Buck, Discrete time Signal Processing 2nd edition Prentice Hall [3] Proakis and Manolakis, Digital Signal Processing Algorithms and applications 3rd edition, chapter 8 pp 617-619; chapter 4 pp 267-269 [4] Digital Signal Processing Lecture series by Prof. Dutta Roy, IITK Lecture 11, NPTEL