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A model of the leakage in the frequency domain and its application to CPA and DPA

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Abstract

This paper introduces a leakage model in the frequency domain to enhance the efficiency of side channel attacks of CMOS circuits. While usual techniques are focused on noise removal around clock harmonics, we show that the actual leakage is not necessary located in those expected bandwidths as experimentally observed by Mateos and Gebotys (A new correlation frequency analysis of the side channel, p 4, 2010). We start by building a theoretical modeling of power consumption and electromagnetic emanations before deriving from it a criterion to guide standard attacks. This criterion is then validated on real experiments, both on FPGA and ASIC, showing an impressive increase of the yield of SCA.

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Appendices

Appendix A: Fourier transform: leakage formula

For a square signal with an amplitude A such as :

$$\begin{aligned} \mathrm{rect}_T (t) = \left\{ \begin{array}{l@{\quad }l} A &{}\mathrm{if}\ t \in \left[ -T/2 , T/2 \right] \\ \mathrm{else} &{} 0 \end{array} \right. \end{aligned}$$
(11)

its Fourier transform is equal to

$$\begin{aligned} \mathrm{FT} \left\{ \mathrm{rect}_T (t) \right\} (f) = AT \mathrm {sinc} (fT) = AT \frac{ \sin (\pi fT)}{\pi fT} \end{aligned}$$
(12)

Moreover, the Fourier transform of a delay \(t_0\) is

$$\begin{aligned} \begin{array}{ll} \mathrm{FT} \left\{ x(t) \right\} (f) = X(f) \\ \mathrm{FT} \left\{ x(t-t_0) \right\} (f) = X(f)e^{-j2 \pi f t_0 } \end{array} \end{aligned}$$
(13)

\(X(f)\) being the Fourier transform of \(x(t)\). From Fig. 2c we can see that the EM signal is equal to the sum of two squares, one of amplitude \(\frac{A}{\alpha T}\) with a delay \(\frac{\alpha T}{2}\) and a period \(\alpha T\) and one of amplitude \(\frac{-A}{(1- \alpha ) T}\) with a delay \(\alpha T + \frac{(1- \alpha ) T}{2}\) and a period \((1- \alpha )T\). Thus, from Eqs. 12, 13 we can deduce that the Fourier transform of the EM model is equal to Eq. 3 :

$$\begin{aligned}&\mathrm{EM}(f) = \frac{A}{\pi f T} \left\{ ~ \frac{1}{\alpha } \mathrm {sin} (\alpha \pi f T) ~ e^{-j \pi f \alpha T }\right. \\&\quad -~ \left. \frac{1}{1-\alpha } \mathrm {sin} \big ((1 - \alpha )\pi f T\big ) ~ e^{-j \pi f \big ((\alpha +1)T\big ) }\right\} \end{aligned}$$

For a function \(g\) and its derivative \({\frac{\mathrm{d}}{\mathrm{d}t}g}\), we have

$$\begin{aligned} \mathrm{TF} \left\{ g \right\} (f) = \frac{\mathrm{TF} \left\{ {\frac{\mathrm{d}}{\mathrm{d}t}g} \right\} (f) }{j2 \pi f} \end{aligned}$$
(14)

Knowing that the EM signal is the derivative of the current signal we can deduce Eq. 2 :

$$\begin{aligned}&\mathrm{POW}(f) = \frac{-jA}{2 \pi ^2 f^2 T}~ \left\{ \frac{1}{\alpha } \mathrm {sin} (\alpha \pi f T) ~ e^{-j \pi f \alpha T }\right. \\&\quad \left. -\frac{1}{1-\alpha } \mathrm {sin} \big ((1 - \alpha )\pi f T\big ) ~ e^{-j \pi f \big ((\alpha +1)T\big ) }\right\} \end{aligned}$$

Appendix B: Fourier transform: leakage repetition formula

$$\begin{aligned} \mathrm{Leakage}(t) = \sum _{k=0}^{k=n}{x\left( t-\frac{k}{F_r}\right) } \end{aligned}$$

Due to the linearity of the Fourier transform and to Eq. 13 of the delay we get Eq. 15 as the Fourier transform of the leakage.

$$\begin{aligned} \mathrm{FT}_L (f) = \sum _{k=0}^{k=n}{ X(f)e^{-j2 \pi f \frac{k}{F_r} } } \end{aligned}$$
(15)

with \( X(f)\!=\! \mathrm{FT} \{ x(t) \} (f) \) and \(\mathrm{FT}_L (f) \!=\! \mathrm{FT}\{\mathrm{Leakage}(t) \}(f) \)

$$\begin{aligned} e^{-jx} = \cos (x) - j \sin (x) \end{aligned}$$
(16)

From Eq. 16, we get:

$$\begin{aligned} \begin{array}{l} \mathrm{FT}_L (f) = X(f) \displaystyle \sum _{k=0}^{k=n}{ \left\{ \cos \left( 2 \pi f \frac{k}{F_r}\right) -j \sin \left( 2 \pi f \frac{k}{F_r}\right) \right\} } \end{array} \end{aligned}$$
$$\begin{aligned}&\left| \mathrm{FT}_L (f) \right| \\&\quad = \left| X(f) \right| \sqrt{ \left( \displaystyle \sum _{k=0}^{k=n}{ \cos \left( 2 \pi f \frac{k}{F_r}\right) } \right) ^2 + \left( \displaystyle \sum _{k=0}^{k=n}{ \sin \left( 2 \pi f \frac{k}{F_r}\right) } \right) ^2 } \end{aligned}$$
$$\begin{aligned} \left( \sum _{k}{a_k} \right) ^2 = \sum _{k}{a_k^2} + 2 \sum _{k}{ \sum _{p>k}{a_k a_p} } \end{aligned}$$
(17)

From Eq. 17 we get:

$$\begin{aligned}&\left| \mathrm{FT}_L (f) \right| \\&\quad \!=\!\left| X(f) \right| \sqrt{~} \bigg \{ \sum _{k=0}^{k=n}{ \cos ^2 \left( 2 \pi f \frac{k}{F_r}\right) } \!+\! \sum _{k=0}^{k=n}{ \sin ^2 \left( 2 \pi f \frac{k}{F_r}\right) } \\&\qquad +\, 2 \sum _{k=0}^{k=n-1}{ \sum _{p=k+1}^{p=n}{ \cos \left( 2 \pi f \frac{k}{F_r}\right) \cos \left( 2 \pi f \frac{p}{F_r}\right) } } \\&\qquad +\, 2 \sum _{k=0}^{k=n-1}{ \sum _{p=k+1}^{p=n}{ \sin \left( 2 \pi f \frac{k}{F_r}\right) \sin \left( 2 \pi f \frac{p}{F_r}\right) } } \bigg \} \end{aligned}$$
$$\begin{aligned} \cos ^2 x + \sin ^2 x = 1 \end{aligned}$$
(18)
$$\begin{aligned}&\cos p \cos q = \frac{1}{2} \left( \cos (p+q) + \cos (p-q) \right) \\&\sin p \sin q = \frac{1}{2} \left( - \cos (p+q) + \cos (p-q) \right) \end{aligned}$$

thus

$$\begin{aligned} \cos p \cos q + \sin p \sin q = \cos (p-q) \end{aligned}$$
(19)

From Eqs. 18, 19 we get :

$$\begin{aligned}&\left| \mathrm{FT}_L (f) \right| \\&\quad = \left| X(f) \right| \sqrt{ (n+1) + 2 \displaystyle \sum _{k=0}^{k=n-1}{ \displaystyle \sum _{p=k+1}^{p=n}{ \cos \left( 2 \pi f \frac{(p-k)}{F_r}\right) } } } \end{aligned}$$
$$\begin{aligned}&\left| \mathrm{FT}_L (f) \right| \\&\quad = \left| X(f) \right| \sqrt{ (n+1) + 2 \displaystyle \sum _{k=1}^{k=n}{ (n-k+1) \cos \left( 2 \pi f \frac{k}{F_r}\right) } } \end{aligned}$$

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Tiran, S., Ordas, S., Teglia, Y. et al. A model of the leakage in the frequency domain and its application to CPA and DPA. J Cryptogr Eng 4, 197–212 (2014). https://doi.org/10.1007/s13389-014-0074-x

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